Limits and continuity[1]

Limits and Continuity

We discuss a number of functions. Our aim is to isolate an important property of a function called continuity. 

1. Let f (x) = sin(x). This is defined for all x    .

[Recall we use radians automatically in order to have the derivative of sin x being cos x.]

2.  Let f (x) = log(x). This is defined for x > 0, and so naturally has a restricted domain. Note also that the domain is an open set.

3. Let f (x) =   when x a, and suppose f (a) = 2a.

4.  Let f (x) = 

5. Let f (x) = 0 if x < 0, and let f (x) = 1 for x  0.

6.  Let f (x) = sin  when x  0 and let f (0) = 0.7. In each case we are trying to study the behaviour of the

function near a particular point. In example 1, the function is well behaved everywhere, there are no problems, and so there is no need to pick out particular points for special care. In example 2, the function is still well behaved wherever it is defined, but we had to restrict the domain. In all of what

follows, we will assume the domain of all of our functions is suitably restricted.

8. We won't spend time in this course discussing standard functions. It is assumed that you know about functions such as sin x, cos x, tan x, log x, exp x, tan-1x and sin-1x, as well as the ``obvious'' ones like polynomials and rational functions -- those functions of the form p(x)/q(x), where p and q are polynomials. In particular, it is assumed that you know these are differentiable everywhere they are defined. We shall see later that this is quite a strong piece of information. In particular, it means they are examples of continuous functions. Note also that even a function like f (x) = 1/x is continuous, because,

wherever it is defined (ie on   - {0}), it is continuous.9. In example 3, the function is not defined at a, but rewriting the

function

10.  = x + a    if x a,    11. we see that as x approaches a, where the function is not

defined, the value of the function approaches 2a. It thus seems very reasonable to extend the definition of f by defining f (a) = 2a. In fact, what we have observed is that

12.  =  (x + a) = 2a.Definition 4.3   Say that f (x) tends to l as x a iff given   > 0,

there is some   > 0 such that whenever 0 < | x - a| <  , then | f (x) - l| <  .

Example :  Let f (x) =   for x 2. Show how to define f (2) in order to make f a continuous function at 2.

13. Solution. We have

 =   = (x2 + 2x + 4)14. Thus f (x) (22 + 2.2 + 4) = 12 as x 2. So defining f (2) =

12 makes f continuous at 2, (and hence for all values of x).15. [Can you work out why this has something to do with the

derivative of f (x) = x3 at the point x = 2?]

One sided limits

Definition   Say that  f (x) = l, or that f has a limit from the left iff given   > 0, there is some   > 0 such that whenever a -   < x < a, then | f (x) - f (a)| <  .

There is a similar definition of ``limit from the right'', writen

as  f (x) = l

Example    Define f (x) as follows:-

f (x) = Calculate the left and right hand limits of f (x) at 2.

Solution. As x 2 -, f (x) = 3 - x 1 +, so the left hand limit is 1. As x 2 +, f (x) = x/2 1 +, so the right hand limit is 1. Thus the left and right hand limits agree (and disagree with f (2), so f is not continuous at 2).

Note our convention: if f (x) 1 and always f (x) 1 as x 2 -, we say that f (x) tends to 1 from above, and write f (x) 1 + etc.

Proposition    If  f (x) exists, then both one sided limts exist and are equal. Conversely, if both one sided limits exits

and are equal, then  f (x) exists.

Proposition   (Continuity Test) The function f is continuous at a iff both one sided limits exits and are equal to f (a).

Example :  Let f (x) =   Show that f is continuous at 1. [In fact f is continuous everywhere].

Solution. We use the above criterion. Note that f (1) = 1. Also

f (x) =  x2 = 1    while     f (x) =  x = 1 = f (1).so f is continuous at 1.

Exercise   Let f (x) =   Show that f is continuous at 0. [In fact f is continuous everywhere]

Example :  Let f (x) = | x|. Then f is continuous in  .

Solution. Note that if x < 0 then | x| = - x and so is continuous, while if x > 0, then | x| = x and so also is continuous. It remains to examine the function at 0. From these identifications, we see

that  | x| = 0 +, while  | x| = 0 +. Since 0 + = 0 - = 0 = | 0|, by the 4.12, | x| is continuous at 0

Results giving Coninuity

Just as for sequences, building continuity directly by calculating limits soon becomes hard work. Instead we give a result that enables us to build new continuous functions from old ones, just as we did for sequences. Note that if f and g are functions andk is a constant, then k.f, f + g, fg and (often) f /g are also functions.

Proposition    Let f and g be continuous at a, and let k be a constant. Then k.f, f + g and fg are continuous at f. Also, if g(a)

 0, then f /g is continuous at a.Proof. We show that f + g is continuous at a. Since, by definition, we have (f + g)(a) = f (a) + g(a), it is enough to show that

(f (x) + g(x)) = f (a) + g(a).

Pick   > 0; then there is some   such that if | x - a| <  ,

then | f (x) - f (a)| <  /2. Similarly there is some   such that

if | x- a| <  , then | g(x) - g(a)| <  /2. Let   = min( , ), and pick x with | x - a| <  . Then

| f (x) + g(x) - (f (a) + g(a))| | f (x) - f (a)| + | g(x) - g(a)| <  /2

+  /2 =  .This gives the result

Note: Just as when dealing with sequences, we need to know that f /g is defined in some neighbourhood of a. This can be shown using a very similar proof to the corresponding result for sequences.

**Proposition    Let f be continuous at a, and let g be continuous at f (a). Then gof is continuous at a

Proof. Pick   > 0. We must find   > 0 such that if | x - a|

<  , then g(f (x)) - g(f (a))| <  . We find   using the given properties of f and g. Since g is continuous at f (a), there is

some   > 0 such that if | y - f (a)| <  , then | g(y) - g(f (a))|

<  . Now use the fact that f is continuous at a, so there is

some   > 0 such that if | x - a| <  , then | f (x) - f (a)| <  . Combining these results gives the required inequality. 

Example :   The function f : x   sin3x is continuous.

Solution. Write g(x) = sin(x) and h(x) = x3. Note that each of g and h are continuous, and that f = goh. Thus f is continuous.

Example:  Let f (x) = tan . Show that f is continuous at every point of its domain.

Solution. Let g(x) =  . Since -1 < g(x) < 1, the function is properly defined for all values of x (whilst tan x is undefined when x = (2k + 1) /2 ), and the quotient is continuous, since

each term is, and since x2 + a2  0 for any x. Thus f is continuous, since f = tanog.

**Example :   Suppose that  sin(1/x) = l; in other words,

assume, to get a contradiction, that the limit exists. Let xn= 1/( n);

then xn  0 as n , and so by assumption, sin(1/xn) = sin(n ) = 0 l as n . Thus, just by looking at a single sequence, we see that the limit (if it exists) can only be l. But instead, consider the

sequence xn = 2/(4n + 1) , so again xn  0 as n . In this

case, sin(1/xn) = sin((4n + 1) /2) = 1, and we must also have l = 1. Thus l does not exist.

Note: Sequences often provide a quick way of demonstrating that a function is not continuous, while, if f is well behaved on each sequence which converges to a, then in fact f is continuous at a. The proof is a little harder than the one we have just given, and is left until next year.

**Infinite limits

There are many more definitions and results about limits. First one that is close to the sequence definition:

Definition    Say that  f (x) = l iff given   > 0, there

is some K such that whenever x > K, then | f (x) - l| <  .

Example :   Evaluate  .

Solution. The idea here should be quite familiar from our sequence work. We use the fact that 1/x  0 as x . Thus

 =      as    x .

LIMITS

Let f and g be two real valued functions with the same domain such that f(x)≤g(x)

For all x in domain of definition, for some a, if both limx→af (x ) ≤

limx→a

g (x)

Identities involving calculus

[ Angle B=θ , A→C as

θ →0 ⇨

CA→0 and

BA→BC as θ →0]

These can be seen from looking at the diagrams.

Sine and angle ratio identity

Proof: Area of ∆ OCD < Area of sector OAC < Area of ∆ OAB

½ r2sinx < ½ r2x < ½ r2tanx [In ∆ OAP, AB = OA tanx]

, so

, so

, or

, so

, but

, so

(By sandwich theorem)

Cosine and angle ratio identity

Proof:

The limits of those three quantities are 1, 0, and 1/2, so the resultant limit is zero.

Cosine and square of angle ratio identity

Proof:

As in the preceding proof,

The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2.

INFORMAL APPROACH TO LIMIT

Consider a function f(x) = x ²−4x−2 .

It is 0/0 form (known as indeterminate form) at x =2, this function is defined ∀ x except x = 2.

If x ≠2, then f(x) = (x−2)(x+2)x−2 = x+2.

The following table exhibits the values of f(x) at points which are close to 2 on its two sides

( left & right on the real line).

x 1.4

1.5

1.6

1.7

1.8

1.9

1.99

2 2.01

2.1

2.2

2.3

2.4

2.5

2.6

f(x)

3.4

3.5

3.6

3.7

3.8

3.9

3.99

0/0

4.01

4.1

4.2

4.3

4.4

4.5

4.6

Graph of f(x)

4

2

-2 0 2

It is evident from the above table and the graph of f(x) that as x increases and comes

closer to 2 from left hand side of 2, the values of f(x) increase and come closer to 4.

x→ 2- , f(x)→4 or , limx→2⁻

f ( x) = 4

From right hand side of 2 , using notation x→ 2+ , f(x)→4 or , limx→2⁺

f ( x) = 4

Question If f(x) = x sin(1/x) , x≠ = 0, x=0 , then test the continuity of f(x) at x=0.

Answer L.H.L. limX→0−O

f (x ) = 0 [∵ -1≤ sin(1/x)≤1 and 0. Finite value =0

] R.H.L. limX→0+O

f (x) =0 ⇨ f(x) is cts.

PROPERTIES OF INFINITY (i) c.∞ → ∞ , if c > 0(ii) c.∞ = 0 , c = 0(iii) c.∞ →-∞ , c < 0.(iv) c∞ = ∞ if c > 1(v) = 0 , 0 ≤ c ≤ 1(vi) = 1 , c = 1.

(vii) limx→ 0−0

1x = -∞ , lim

x→ 0+0

1x = ∞.

INTERDETERMINATE FORMSEXAMPLES:

00 ( lim

X→ 1

X ²−1X−1 )

∞∞ ( lim

X→∞

X ²+1X ³+2 )

0.∞ ( limX→π /2

¿)

∞−∞ ( limX→1

1X ²−1 - 2

X ⁴−1 )

1∞ ( limX→π /2

¿ )00 ( lim

X→0¿¿¿ )

∞0 ( limX→ 0¿¿

¿

o PROBLEM 3 : Determine if the following function is continuous at x=0 .

SOLUTION : Function f is defined at x=0 since

i.) f(0) = 2 .

The left-hand limit

= 2 .

The right-hand limit

= 2 .

Thus, exists with

ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 .

SOLUTION : Function h is not defined at x=-1 since it leads to division by zero. Thus,

i.) h(-1)

does not exist, condition i.) is violated, and function h is NOT continuous at x = -1 .

o

PROBLEM 5 : Check the following function for continuity at x=3 and x=-3 .

SOLUTION 5 : First, check for continuity at x=3 . Function f is defined at x=3 since

i.) .

The limit

(Circumvent this indeterminate form by factoring the numerator and the denominator.)

(Recall that A2 - B2 = (A-B)(A+B) and A3 - B3 = (A-B)(A2+AB+B2 ) . )

(Divide out a factor of (x-3) . )

=

,

i.e.,

ii.) .

Since,

iii.) ,

all three conditions are satisfied, and f is continuous at x=3 . Now, check for continuity at x=-3 . Function f is not defined at x = -3 because of division by zero. Thus,

i.) f(-3)

does not exist, condition i.) is violated, and f is NOT continuous at x=-3 .

o PROBLEM 11 : For what values of x is the following function continuous ?

SOLUTION 11 : Consider separately the three component functions

which determine f . Function is continuous for x > 1 since it is the quotient of continuous functions and the denominator is never

zero. Function y = 5 -3x is continuous for since it is a

polynomial. Function is continuous for x < -2 since it is the quotient of continuous functions and the denominator is never zero. Now check for continuity of f where the three components are joined together, i.e., check for continuity at x=1 and x=-2 . For x = 1 function f is defined since

i.) f(1) = 5 - 3(1) = 2 .

The right-hand limit

=

(Circumvent this indeterminate form one of two ways. Either factor the numerator as the difference of squares, or multiply by the conjugate of the denominator over itself.)

= 2 .

The left-hand limit

=

= 5 - 3(1)

= 2 .

Thus,

ii.) .

Since

iii.) ,

all three conditions are satisfied, and function f is continuous at x=1 . Now check for continuity at x=-2 . Function f is defined at x=-2 since

i.) f(-2) = 5 - 3(-2) = 11 .

The right-hand limit

=

= 5 - 3( -2)

= 11 .

The left-hand limit

=

= -1 .

Since the left- and right-hand limits are different,

ii.) does NOT exist,

condition ii.) is violated, and function f is NOT continuous at x=-2 . Summarizing, function f is continuous for all values of x EXCEPT x=-2 .

o PROBLEM 13 : Determine all values of the constants A and B so that the following function is continuous for all values of x .

SOLUTION 13 : First, consider separately the three components which

determine function f . Function y = Ax - B is continuous for for any values of A and B since it is a polynomial. Function y = 2x2 + 3Ax +

B is continuous for for any values of A and B since it is a polynomial. Function y = 4 is continuous for x > 1 since it is a polynomial. Now determine A and B so that function f is continuous at x= -1 and x= 1 . First, consider continuity at x= -1 . Function f must be defined at x= -1 , so

i.) f(-1)= A(-1) - B = - A - B .

The left-hand limit

=

= A (-1) - B

= - A - B .

The right-hand limit

=

= 2(-1)2 + 3A(-1) + B

= 2 - 3A + B .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

so that

2A - 2B = 2 ,

or

(Equation 1)

A - B = 1 .

Now consider continuity at x=1 . Function f must be defined at x=1 , so

i.) f(1)= 2(1)2 + 3A(1) + B = 2 + 3A + B .

The left-hand limit

=

= 2(1)2 + 3A(1) + B

= 2 + 3A + B .

The right-hand limit

=

= 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

or

(Equation 2)

3A + B = 2 .

Now solve Equations 1 and 2 simultaneously. Thus,

A - B = 1 and 3A + B = 2

are equivalent to

A = B + 1 and 3A + B = 2 .

Use the first equation to substitute into the second, getting

3 (B + 1 ) + B = 2 ,

3 B + 3 + B = 2 ,

and

4 B = -1 .

Thus,

and

.

For this choice of A and B it can easily be shown that

iii.)

and

iii.) ,

so that all three conditions are satisfied at both x=1 and x=-1 , and function f is continuous at both x=1 and x=-1 . Therefore, function f is

continuous for all values of x if and .

o PROBLEM 14 : Show that the following function is continuous for all values of x .

SOLUTION 14 : First describe f using functional composition. Let g(x) = -1/x2 and h(x) = ex . Function h is well-known to be continuous for all values of x . Function g is the quotient of functions continuous for all values of x , and is therefore continuous for all values of x except x=0 , that x which makes the denominator zero. Thus, for all values of x except x=0 ,

f(x) = h ( g(x) ) = e g(x) = e -1/x2

is a continuous function (the functional composition of continuous functions). Now check for continuity of f at x=0 . Function f is defined at x=0 since

i.) f(0) = 0 .

The limit

(The numerator approaches -1 and the denominator is a positive number approaching zero.)

,

so that

= 0 ,

i.e.,

ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 . Thus, f is continuous for all values of x .

o PROBLEM 15 : Let

Show that f is continuous for all values of x .

SOLUTION 15 : First show that f is continuous for all values of x .

Describe f using functional composition. Let , , and k(x) = x2 . Function h is well-known to be continuous for all values of x . Function k is a polynomial and is therefore continuous for all values of x . Function g is the quotient of functions continuous for all values of x , and is therefore continuous for all values of x except x=0 , that x which makes the denominator zero. Thus, for all values of x except x=0 ,

is a continuous function (the product and functional composition of continuous functions). Now check for continuity of f at x=0 . Function f is defined at x=0 since

i.) f(0) = 0 .

The limit does not exist since the values of oscillate

between -1 and +1 as x approaches zero. However, for

so that

.

Since

,

it follows from the Squeeze Principle that

ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 . Thus, f is continuous for all values of x .

o PROBLEM 1 : Compute .

SOLUTION1 : Note that DOES NOT EXIST since values of

oscillate between -1 and +1 as x approaches 0 from the left.

However, this does NOT necessarily mean that does not exist ! ? #. Indeed, x3 < 0 and

for x < 0. Multiply each component by x3, reversing the inequalities and getting

or

.

Since

,

it follows from the Squeeze Principle that

.

o PROBLEM 2 : Compute .

SOLUTION2 : First note that

,

so that

and

.

Since we are computing the limit as x goes to infinity, it is reasonable to assume that x+100 > 0. Thus, dividing by x+100 and multiplying by x2, we get

and

.

Then

=

=

=

= .

Similarly,

= .

Thus, it follows from the Squeeze Principle that

= (does not exist).

DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS

None of the six basic trigonometry functions is a one-to-one function. However, in the following list, each trigonometry function is listed with an appropriately restricted domain, which makes it one-to-one.

1. for

2. for

3. for

4. for , except

5. for , except x = 0

6. for

Because each of the above-listed functions is one-to-one, each has an inverse function. The corresponding inverse functions are

1. for

2. for

3. for

4. arc for , except

5. arc for , except y = 0

6. arc for

In the following discussion and solutions the derivative of a function

h(x) will be denoted by or h'(x) . The derivatives of the above-mentioned inverse trigonometric functions follow from trigonometry identities, implicit differentiation, and the chain rule. They are as follows.

1.

2.

3.

4. arc

5. arc

6. arc

Important points: (1)when limx→cf (x ) exists but either f© does not exist

or limx→cf (x ) ≠f(c), we say that f has a reomovable discts. ;otherwise f

has non- reomovable discts.(2) composition of real valued fns. of f and g (fog) is defined at c. if g is cts. At c and if f is cts. At g© then fog is cts. At c, fog(x)=f(g(x)) is defined whenever the range of g is subset of domain of f. [Ex. sin(x²)]Some questions on differentiability:1. Show that f(x)=x2 is diff. at x=1, find f’(1)2. Prove that f(x) = [x], 0<x<3 is not diff. at x=1 & 2.

3. Discuss the continuity & diff. of f(x)={x ² sin ( 1x ) , x≠0

0 , x=0

4. Show that f(x)={1+x , if x≤25−x ,if x>2 is not diff. at x=2.

5.For which value of a &b the function f(x) = {x ²+2x , x ≤0ax+b , x>0 is diff. at x=0

Solutions:

1. R f’(1) = limh→ 0

f (1+h )−f (1)h

= limh→0

h ²+2hh =2

L f’(1)= limh→0

f (1−h )−f (1)−h

=limh→ 0

h ²−2h−h =2 ⇨ L.H.D = R.H.D=2

2. R f’(1) = limh→0

f (1+h )−f (1)h

= limh→0

[ 1+h ]−[1]h =1-1/h=0{∵ [1+h]=1}

L f’(1)= limh→0

f (1−h )−f (1)−h

=limh→0

[ 1−h ]−[1]−h = 0-1/-h=∞ ⇨R.H.D ≠L.H.D ⇨ f is

not diff. at x=1. At x=2, similarly f is not diff. at x=2, ∵ R.H.D= 0≠ L.H.D=∞ {[2+h]=2 & [2-h] = 1}3. L.H.lt limh→0

¿ =0 = R.H.lt ⇨ f is cts. At x=0

L.H.D limh→ 0

f (−h )−f (0)−h

= limh→0

(−h ) sin( 1−h

) =0= R.H.D ⇨ diff.

4. R.H.D. limh→0

f (2+h )−f (2)h

= limh→0

5−(2+h )−(1+2)h

= -1

L.H.D limh→ 0

f (2−h )−f (2)−h

= limh→ 0

[1+(2−h ) ]−[1+2 ]−h

= 1 ⇨ not diff.

5. Given L.H.D= R.H.D ⇨ limh→ 0

f (0−h )−f (0)−h

= limh→ 0

f (0+h )−f (0)h

limh→ 0

[ (−h )2+2 (−h ) ]−[ 0+2×0 ] /−h =limh→0

ah+b−0h

a=2 & b=0 [ ∵ f is cts. ⇨b=0]

ASSIGNMENT(continuity & differentiability) (XII)

**Question 1 Determine a and b so that the function f given by

f(x) = 1−sin ² x3cos ² x , x<п/2

=a, x=п/2

= b(1−sinx)

(п−2x ) ² , x>п/2

Is continuous at x=п/2.

Answer [a = 1/3 , b = 8/3]

**Question 2 Find k such that following functions are continuous at indicated point

(i) f(x) ={1−cos4 x8x ²

, x ≠0

k , x=0 at x=0

(ii) f(x) = (2x+2 - 16)/(4x – 16) , x≠2

= k, x = 0 at x=2.

Answer [ (i) k=1,(ii) k=1/2]

**Question 3 The function f is defined as {x ²+ax+b ,0≤x<23 x+2 ,2≤ x≤4

2ax+5b ,4< x≤8

If f(x) is continuous on [0,8], find the values of a and b.

Answer [a=3,b=-2]

** Question 4 If f(x) = {√1+ px−√1−pxx

,−1≤ x<0

2 x+1x−1

,0≤ x ≤1 is continuous in the [-

1,1], find p.

Answer [p=-1]

**Question 5 Find the value of a and b such that the f(x) defined as

f(x) = { x+a√2 sinx ,0≤ x<п/42 xcotx+b ,п /4≤ x≤п/2

acos2 x−bsinx ,п2<x≤п

is continuous for all values of x in

[0,п].

ANSWER [a=п/6 , b=-п/12]

** Question 6 Prove that limx→π /4

tan3 x−tanx

cos (x+ √π4

) = -4

[ Hint: Nr. Can be written as tanx(tanx-1)(tanx+1) =- [tanx(cosx-sinx)(tanx+1)]/cosx Cosx-sinx = √ 2 cos¿) ]

**Question 7 Prove that (i) limx→ 1

√ 2

x−cos (sin−1 x )1−tan (sin−1 x ) = −1

√2 [ Hint: put x=

sin ]Ѳ

(ii) limx→∞x ¿¿ ) = -3/2. [Hint: π4 = tan−11 & use formula of tan−1 x−tan−1 y ]

Question 8 f(x) = a x2+b

x2+1 , limx→0

f (x ) =1 & limx→∞f (x) =1, then p.t. f(-2)=f(2)=1.

[ Hint: limx→∞

1

x2 =0]

Question 9 limx→0

ex−1√1−cosx

[Dr. = √2|sinx/2| &limx→0

e x−1x

=1

|sinx/2| =+ve & -ve as x→0+ & x→0- , ⇨ limit does not exist] Question 10 Show that the function

f(x)¿ {sin 3 xtan 3 x

, x<0

32, x=0

log(1+3 x)e2x−1

, x>0

is continuous at x=0.

[Hint: use limx→0

e x−1x

=1 , limx→0

log (1+x)x =1]

Question11 Show that f(x) = |x-3|,x∊R is cts. But not diff. at x=3.

[Hint:show L.H.lt=R.H.lt by |x-3| = x-3, if x ≥3 and –x+3, if x<3, L.hd=-1≠1(R.h.d)

QUESTION 12 Discuss the continuity of the fn. f(x) = |x+1|+|x+2|, at x

= -1 & -2 [Hint:f(x) = {−2x−3 ,when x←21 ,when−2≤x←1

2 x+3 ,when x ≥−1 yes cts. At x=-1,-2

Question 13 Find the values of p and q so that f(x) ={x ²+3x+ p , if x≤12x+2 , if x>1 is

diff. at x = 1. [ answer is p=3 , q=5]

Question 14 For what choice of a, b, c if any , does the function

F(x) = {ax ²+bx+c ,0≤ x≤1bx−c ,1<x ≤2

c , x>2 becomes diff at x=1,2 & show that a=b=c=0.

Question15For what values a,b f(x)={ e2x−1 ,when x ≤0

ax+ bx ²2

,when x>0 is diff.at x=0

[Hint: L.H.d= 2by using limx→0

ex−1x

=1& R.H.d=a, since f‘(x)=0exists, a=2,b∊R]

ASSIGNMENT WITH HINTS (XI)

Question.1 Evaluate

** limx→0

1−cosx √cos 2xx ²

[Hint limx→0

1−cosx √cos 2xx ²

1+cosx √cos2 x1+cosx √cos2 x

limx→0

1− (1−sin2 x )(1−2 si n2 x )x ² ¿¿

¿]

Question.2 limx→0

1−cosxcos2xcos 3xsin ²2 x

[Hint limx→0

1−1/4 (1+cos2 x+cos4 x+cos 6 x)sin ²2 x =7/4]

Question.3 limx→π

sin3 x−3 sinx(π−x) ³ [use sin3x=3sinx-4sin³x, put (π−x)

=h ,answer is -4]

Question.4 limx→π /6

√3 sinx−cosxx−π /6

[Hint limx→π /6

2( √32sinx−1

2cosx)

x−π /6

limx→π /6

2sin (x−π6)

(x−π /6) = 2]

Question.5 limx→0

tanx−sinxx ³ [hint: put tanx=sinx/cosx, answer is 1/2]

Question.6 limx→0

sinx−2sin 3 x+sin 5 xx

[Hint: use sinx-siny=2cos(x+y)/2.sin(x-y)/2 , answer is 0]

Question.7 limx→π /2

cotx−cosxcos ³ x [ answer is ½]

Question.8 limx→0

1−cosxsin ²2 x [ answer is 1/8]

Question.9 Let f(x) = { 3−x ² , x≤−2ax+b ,−2< x<2

x2

2, x ≥2

Find a,b so that limx→2f (x ) and lim

x→−2f (x) exist.

[Hint: limx→2f (x ) exists lim

x→2⁻f ( x )=¿ limx→2

¿ (ax+b)= limx→2+¿ f (x)¿

¿ =limx→2

¿ x2/2 2a+b=2……(1)

Similarly -2a+b = -1……(2)

a=3/4,b=1/2]

Question.10 lim

x→3+¿x

[x]¿

¿ and

limx→3−¿

x[x ]

¿

¿ where [x] denotes the integral part

of x. Are they equal?

[HINT: lim

x→3+¿x

[x]¿

¿ = 3/3=1,x>3

limx→3−¿

x[x ]

¿

¿ = 3/2 ,x<3]

Question.11 limx→5−0

[−x ] and limx→5+0

[−x], are they equal?

-5 as -x→−5 -6

**Question.12 Test the continuity of the function at x=0, f(x) = e1x

1+e1x

,

x ≠0

0 , x=0 [Hint e-∞ =0,e∞ =∞]

Derivatives (XI)

BY FIRST PRINCIPLE

o Use the limit definition to compute the derivative, f'(x), for

.

SOLUTION :

(Get a common denominator for the expression in the numerator.

Recall that division by is the same as multiplication by . )

(Algebraically and arithmetically simplify the expression in the numerator. It is important to note that the denominator of this

expression should be left in factored form so that the term can be easily eliminated later.)

(The term now divides out and the limit can be calculated.)

o Use the limit definition to compute the derivative, f'(x), for

SOLUTION 6 :

(Recall a well-known trigonometry identity :

.)

(Recall the following two well-known trigonometry limits :

and .)

SOLUTION 3 : Differentiate . Apply the quotient rule.

Then

(Recall the well-known trigonometry identity .)

.

SOLUTION 4 : Differentiate . Apply the product rule.

Then

SOLUTION 5 : Differentiate . This is NOT a product of functions. It's a composition of functions. Apply the chain rule. Then

.

SOLUTION 6 : Differentiate . Apply the product rule first, followed by the chain rule. Then

(Ncert)limits & derivatives (xi)

Question 4:

Evaluate the Given limit:

Question 10:

Evaluate the Given limit:

Question 12:

Evaluate the Given limit:

Question 14:

Evaluate the Given limit:

Question 15:

Evaluate the Given limit:

Question 16:

Evaluate the given limit:

Question 17:

Evaluate the Given limit:

Question 18:

Evaluate the Given limit:

Question 20:

Evaluate the Given limit:

Question 21:

Evaluate the Given limit:

Question 22:

Question 23:

Find f(x) and f(x), where f(x) =

Question 24:

Find f(x), where f(x) =

Question 25:

Evaluate f(x), where f(x) =

Question 26:

Find f(x), where f(x) =

Question 28:

Suppose f(x) = and if f(x) = f(1) what are possible values of a and b?

Question 30:

If f(x) = .

For what value (s) of a does f(x) exists?

Question 31:

If the function f(x) satisfies , evaluate.

Question 32:

If . For what integers m and n does

and exist?

ANSWERS: 4. 19/2, 10. 2, 12. -1/4, 14. A/B, 15. 1/п, 16. 1/п, 17. 4, 18. (a+1)/b, 20. 1, 21. 0, 22. 2, 23. 3,6, 24. Limit does not exist at x=1, 25. Limit does not exist at x=0, 26.

Limit does not exist at x=0, 28. A=0,b=4, 30. f(x)

exists for all a≠0, 31. 2, 32. we need m=n; and

exists for any integral value of m & n.

DERIVATIVES

Question 1:

Find the derivative of the following functions from first principle:

(i) –x (ii) (–x)–1 (iii) sin (x + 1)

(iv)

Solution of (iv) : use formula of first principle(ab-initio)

We have f’(x) = limh→0

f (x+h )−f ( x )h = lim

h→0

cos (x+h−п8)−cos (x−п

8)

h =

limh→0

−2sin( x−п8+h) . sin( h2 )h

= sin(x-п/8) [ limh→0

sin ( h2)

h /2=1

, cosA-cosB= -

2sin(A+B)/2.sin(A-B)/2]

Question 11:

Find the derivative of the following functions:

(i) sin x cos x (ii) sec x (iii) 5 sec x + 4 cos x

(iv) cosec x (v) 3cot x + 5cosec x

(vi) 5sin x – 6cos x + 7 (vii) 2tan x – 7sec x

Solution of (i) (2 sin x cos x)/2= sin2x/2 , derivative is ½ (2cos2x) = cos2x or can use Leibnitz product rule.

Question 16:

Find the derivative of the following functions :

Solution: by quotient rule or take derivative after simplification

It can be written as sin ( π

2−x )

1+cos(π2−x)

= 2sin ( π4 −x ) .cos ( π

4−x )

2co s2( π4 −x) =tan(

( π4 −x)Derivative is - sec2(

π4−x ¿.

Question 17:

Find the derivative of the following functions:

Solution: multiple & divide by cosx, we get tanx+1tanx−1 = -

tan(п/4+x) Derivative is - sec2(п/4+x) [simplest method]

Question 18:

Find the derivative of the following functions

Solution: can be written as 1−cosx1+cosx = tan2(x/2)

Derivative is 2 tan(x/2) . sec2(x/2) . ½ = tan(x/2) . sec2(x/2).

Question 23:

Find the derivative of the following functions (x2 + 1) cos x

Question 24:

Find the derivative of the following functions: (ax2 + sin x) (p + q cos x)

Question 25:

Find the derivative of the following function

Question 26:

Find the derivative of the following function :

Question 27:

Find the derivative of the following function:

Question 28:

Find the derivative of the following functions :

Question 29:

Find the derivative of the following functions : (x + sec x) (x – tan x)

Question 30:

Find the derivative of the following functions: