Selected Problems on Limits and Continuity
Jan 15, 2016
Selected Problems on Limits and Continuity
1. lim 2x3 – 5x2 + 3xx1 3x2 – 5x + 2
= lim x(2x2 – 5x + 3) x 1 3x2 – 5x + 2
= lim x (2x – 3)(x – 1)(3x – 2)(x – 1)x 1
= (1) ( 2 (1) – 3)
( 3(1) – 2 )= (1)(2 – 3 )
(3 – 2 ) - 1
2. lim x – 1
x2 + 3 - 2 x1● x2 + 3 + 2
x2 + 3 + 2
= lim (x – 1) ( x2 + 3 + 2)x1 x2 + 3 − 4
= lim (x – 1) ( x2 + 3 + 2 )x1 x2 – 1
= lim (x – 1) ( x2 + 3 + 2 )x1 (x + 1)(x – 1 )
= lim ( x2 + 3 + 2 )x1 x + 1
= 12 + 3 + 2 1 + 1
= 4 + 2
2
= 2
3. lim 2x tan 3xx 0
= lim 2x x 0 sin 3x
cos 3x
= lim 2x x 0
● cos 3xsin 3x
1
= lim 2x cos 3x x 0 sin 3x
= lim 2 x 0
● xsin 3x
● cos 3x
= lim 2 x 0
●
sin 3x● cos 3xx ● 3
3
= lim 2 x 0
●
sin 3x
3x ● cos 3x● 13
= lim 2 ● 1 ●x 0
13
● cos 3x
= 2 ( 1 ) ( 1 ) cos 3(0)3
= 2 ● cos 03
= 2 ● 1 3
2/3
lim x4 – a4
xa x2 – a2
= lim (x2 + a2) (x2 – a2)x a
x2 – a2
= lim (x2 + a2) x a
= a2 + a2 2a2
4.
5. Find the value of a and b such that
lim a + bx - 3 x 0 x
= 3
lim a + bx - 3 x 0 x
● a + bx + 3
a + bx + 3
lim a + bx - 3 x 0 (x) ( a + bx + 3 )
3
lim 3 + bx - 3x 0 (x) ( 3 + bx + 3 )
= 3
= 3
lim bxx 0 (x) ( 3 + bx + 3 )
lim bx 0 ( 3 + bx + 3 )
b
( 3 + b(0) + 3 )
= 3
= 3
= 3
b
3 + 3
b2 3
= 31
b = 2 9
b = 6 a = 3
= 3
6. The function below is discontinuous at x = 3. Redefine this function to make it continuous.
f (x) = 2x2 – 7x + 3 x – 3
Note: To make a function continuous, we are essentially going to fill up the hole.
f (x) = 2x2 – 7x + 3 x – 3
f(x) = (2x – 1 )(x – 3 )
x – 3 f(x) = 2x – 1
x – 3 = 0
x = 3
f(3) = 2(3) – 1
f(3) = 5
hole: (3, 5)
f (x) = 2x2 – 7x + 3 x – 3
This function is discontinuous at (3,5)
f(x) = 2x2 – 7x + 3 if x 3
5 if x = 3
x – 3
The graph of these two functions are identical except for the hole.
7. The function below is continuous. Find the value of the constant c.
f(x) = x2 – c2 if x < 4
cx + 20 if x 4
Since each part of these function is a polynomial function, then each of them is continuous. Thus the only possible point of discontinuity is at the boundary point.
lim x2 – c2
42 – c2
lim cx + 20
c(4) + 2042 – c2 = 4c + 2016 – c2 = 4c + 20
16 – c2 – 4c – 20 = 0- c2 – 4c – 4 = 0c2 + 4c + 4 = 0 (c + 2)(c + 2) = 0
c = - 2
x4 x4
8. Find the values of a and b such if the function f below is continuous.
f(x) = ax2 + x – b if x 2
2x + b if 2 < x < 5
2ax – 7 if x 5
a(2)2 + 2 – b 2(2) + b
4a + 2 – b 4 + b
4a + 2 – b = 4 + b
4a + 2 – b – 4 – b = 0
4a – 2b - 2 = 0
lim ax2 + x – b lim 2x + b x2 x2
8. Find the values of a and b such if the function f below is continuous.
f(x) = ax2 + x – b if x 2
2x + b if 2 < x < 5
2ax – 7 if x 5
lim 2ax - 7lim 2x + b
2(5) + b 2a(5) – 7
10 + b = 10a – 7
b = 10a – 7 – 10
b = 10a - 17
2a – 2b - 2 = 0
4a – 2 (10a – 17 ) - 2 = 0
x5 x5
4a – 20a + 34 - 2 = 0
- 16a + 32 = 0
- 16a = - 32a = - 32
- 16a = 2
b = 10a - 17
b = 10(2) – 17
b = 20 – 17
b = 3
9. Find the value of a such that the following is continuous.
f(x) = ax if x 0tan x
x2 – 2 if x < 0
lim axx0 tan x
=lim x2 - 2
x0
? = - 2
lim ax x 0 tan x
axsin xcos x
ax ● cos x sin x
ax cos x sin x
lim x0
lim x0
=
lim x0
lim x0
a ● x ● cos x sin x
lim x0
a ● 1 ● cos x
= a ( 1 ) ( cos 0)
= a ( 1 ) ( 1 )
= a
lim axx0 tan x
= lim a2 - 2x0
? = - 2
a = - 2
10. Given that lim f(x) = 5 x3
lim g(x) = 0x3
lim h(x) = - 8x3
find the following;
a) lim ( f(x) + g(x) )x3
= lim f(x) + lim g(x) x3 x3
= 5 + 0 = 5
10. Given that lim f(x) = 5 x3
lim g(x) = 0x3
lim h(x) = - 8x3
find the following;
b) lim ( f(x) / h(x) )x3
= lim f(x) / lim h(x) x3 x3
= 5/-8
10. Given that lim f(x) = 5 x3
lim g(x) = 0x3
lim h(x) = - 8x3
find the following;c) lim 2 h(x)
x3 f(x) – h(x)
= 2 lim h(x)x3
lim f(x) - lim h(x)x3 x3
= 2 ( - 8 )
5 – ( - 8 )
= - 1613
10. Given that lim f(x) = 5 x3
lim g(x) = 0x3
lim h(x) = - 8x3
find the following;
d) lim x2 f(x)x3
= lim x2 x3 x3
= 32 ● 5 = 45
● lim f (x )