Lecture Notes 2 Calc I: Limits, Continuity and Derivatives, including exercises. What do we need limits for? Two motivating examples: 1. The velocity problem: Let’s suppose that a bomb is falling down from a bomber airplane at a height of 490 meters at time t = 0: the height h(t) is a function of the time: () = − . What is the velocity, in meters per second, of the bomb at time t = 1? The heights after 1, 1.5, 2 and 3 seconds are: ℎ(1) = 485.1 , ℎ(1.5) = 478.975 , ℎ(2) = 470.4 and ℎ(3) = 445.9 During the time interval [1, 3] the bomb is falling 485.1 - 445.9 = 39.2 m, so 19.6 meter per second: if the bomb would fall at a constant velocity on the interval [1, 3], the velocity would be –19.6 m/s. This can be interpreted as the slope of the line through the points (1, 485.1) and (3, 445.9) But we know that the velocity increases as t increases: For the interval [1, 2] the constant velocity would be slope = ℎ(2)−ℎ(1) 2−1 = 470.4−485.1 2−1 = 14.7 / For the interval [1, 1.5] the constant velocity would be = ℎ(1.5)−ℎ(1) 1.5−1 = 478.975−485.1 1.5−1 = 12.25 / Clearly, by choosing the interval [1, a] we find similarly = ℎ()−ℎ(1) −1 / on this interval. The velocity after exactly 1 second can be approximated by choosing a very close to 1: this process is called taking the limit of ℎ()−ℎ(1) −1 for a approaching 1: lim →1 ℎ()−ℎ(1) −1 is the velocity at t = 1. 2. The tangent problem: Consider the function = () = 2 and the point (1, 1) of its graph. Through this point we can draw the tangent: the line is “touching” the graph in this point and its slope gives the “direction” of the graph at x = 1. If we consider the line through (1, 1) and a point (x, x 2 ), close to (1, 1) on the graph, but not equal to (1,1): the slope of this line is ()−(1) −1 = 2 −1 −1 , so the slope of the tangent is lim →1 2 −1 −1 A first computation of a limit: how to compute the limit of the tangent problem: lim →1 2 −1 −1 The function 2 −1 −1 is defined for all x but not for x = 1: the domain is ℝ\{1}. Substituting x = 1 in the function we get a computation of the form: 0 0 , which is indeterminate. Nevertheless for all x ≠ 1we can simplify the function 2 −1 −1 = (+1)(−1) −1 =+1, whose graph is a straight line y = x + 1, except for the point (1, 2), as the function is not defined for x = 1. As x approaches 1 very closely we will approach the point (1, 2) very closely, so the value of 2 −1 −1 will be very close to 2. This reasoning above can be summarized using the limit notation: → − − = → ( + )( − ) − = → ( + ) = Note that substituting x = 1 in 2 −1 −1 leads to the (indeterminate) result 0 0 , but substituting x =1 in +1 leads to 2. And we can cancel −1 in (+)(−) − since x → 1 (x approaches 1) implies x ≠ 1. The “approach” of x to 1 can be done either from the right / from above (notation ↓ → + ) or from the left / from below (notation ↑ → − ) The reasoning above remains valid for this one sided approach: lim ↓1 2 −1 −1 = lim ↓1 ( + 1)( − 1) −1 = lim ↓1 ( + 1) = 2 and lim ↑1 2 −1 −1 =2 Note that if x approaches 1 from the right it means that x > 1. Similarly, if ↑1 implies x <1. In this case → () = 2 for x approaches 1 exists, confirmed by the fact that the limits from the right and the left exist and both are equal to the same result: lim ↓1 () = lim ↑1 () = 2
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Lecture Notes 2 Calc I: Limits, Continuity and Derivatives, including exercises.
What do we need limits for? Two motivating examples:
1. The velocity problem: Let’s suppose that a bomb is falling down from a bomber airplane at a height of
490 meters at time t = 0: the height h(t) is a function of the time: 𝒉(𝒕) = 𝟒𝟗𝟎 − 𝟒. 𝟗𝒕𝟐
What is the velocity, in meters per second, of the bomb at time t = 1?
The heights after 1, 1.5, 2 and 3 seconds are: ℎ(1) = 485.1 𝑚, ℎ(1.5) = 478.975 𝑚,
ℎ(2) = 470.4 𝑚 and ℎ(3) = 445.9 𝑚
During the time interval [1, 3] the bomb is falling 485.1 - 445.9 = 39.2 m, so 19.6 meter per second: if
the bomb would fall at a constant velocity on the interval [1, 3], the velocity would be –19.6 m/s.
This can be interpreted as the slope of the line through the points (1, 485.1) and (3, 445.9)
But we know that the velocity increases as t increases:
For the interval [1, 2] the constant velocity would be slope 𝑣 =ℎ(2)−ℎ(1)
2−1=
470.4−485.1
2−1= 14.7 𝑚/𝑠
For the interval [1, 1.5] the constant velocity would be 𝑣 =ℎ(1.5)−ℎ(1)
1.5−1=
478.975−485.1
1.5−1= 12.25 𝑚/𝑠
Clearly, by choosing the interval [1, a] we find similarly 𝑣 =ℎ(𝑎)−ℎ(1)
𝑎−1 𝑚/𝑠 on this interval.
The velocity after exactly 1 second can be approximated by choosing a very close to 1: this process is
called taking the limit of ℎ(𝑎)−ℎ(1)
𝑎−1 for a approaching 1: lim
𝑎→1
ℎ(𝑎)−ℎ(1)
𝑎−1is the velocity at t = 1.
2. The tangent problem: Consider the function 𝑦 = 𝑓(𝑥) = 𝑥2 and the point (1, 1) of its graph.
Through this point we can draw the tangent: the line is “touching” the graph in this point and its slope
gives the “direction” of the graph at x = 1.
If we consider the line through (1, 1) and a point (x, x2), close to (1, 1) on the graph, but not equal to
(1,1): the slope of this line is 𝑓(𝑥)−𝑓(1)
𝑥−1=
𝑥2−1
𝑥−1, so the slope of the tangent is lim
𝑥→1 𝑥2−1
𝑥−1
A first computation of a limit: how to compute the limit of the tangent problem: lim𝑥→1
𝑥2−1
𝑥−1
The function 𝑥2−1
𝑥−1 is defined for all x but not for x = 1: the domain is ℝ\{1}.
Substituting x = 1 in the function we get a computation of the form: 0
0 , which is indeterminate.
Nevertheless for all x ≠ 1we can simplify the function 𝑥2−1
𝑥−1=
(𝑥+1)(𝑥−1)
𝑥−1= 𝑥 + 1, whose graph is a
straight line y = x + 1, except for the point (1, 2), as the function is not defined for x = 1.
As x approaches 1 very closely we will approach the point (1, 2) very closely, so the value of 𝑥2−1
𝑥−1
will be very close to 2.
This reasoning above can be summarized using the limit notation:
𝐥𝐢𝐦𝒙→𝟏
𝒙𝟐 − 𝟏
𝒙 − 𝟏= 𝐥𝐢𝐦
𝒙→𝟏
(𝒙 + 𝟏)(𝒙 − 𝟏)
𝒙 − 𝟏= 𝐥𝐢𝐦
𝒙→𝟏(𝒙 + 𝟏) = 𝟐
Note that substituting x = 1 in 𝑥2−1
𝑥−1 leads to the (indeterminate) result
0
0 , but substituting x =1 in
𝑥 + 1 leads to 2. And we can cancel 𝑥 − 1 in (𝒙+𝟏)(𝒙−𝟏)
𝒙−𝟏 since x → 1 (x approaches 1) implies x ≠ 1.
The “approach” of x to 1 can be done either from the right / from above (notation 𝒙 ↓ 𝟏 𝑜𝑟 𝒙 → 𝟏+) or from the left / from below (notation 𝒙 ↑ 𝟏 𝑜𝑟 𝒙 → 𝟏−) The reasoning above remains valid for this one sided approach:
lim𝑥↓1
𝑥2 − 1
𝑥 − 1= lim
𝑥↓1
(𝑥 + 1)(𝑥 − 1)
𝑥 − 1= lim
𝑥↓1(𝑥 + 1) = 2 and 𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 lim
𝑥↑1
𝑥2 − 1
𝑥 − 1= 2
Note that if x approaches 1 from the right it means that x > 1. Similarly, if 𝑥 ↑ 1 implies x <1.
In this case 𝐥𝐢𝐦𝒙→𝟏
𝒇(𝒙) = 2 for x approaches 1 exists, confirmed by the fact that the limits from the
right and the left exist and both are equal to the same result: lim𝑥↓1 𝑓(𝑥) = lim
𝑥↑1 𝑓(𝑥) = 2
Infinite limits can occur if substitution of x = a leads to an expression of the form 𝑐
0 where c ≠ 0, e.g.:
Consider the limit at 2 of the function 𝑓(𝑥) =1
(𝑥−2)2: substituting x = 2, we find
1
0 .
Applying the limits from the right and from the left (if x is near 2 we divide 1 by a small positive number):
lim𝑥↓2
1
(𝑥 − 2)2= +∞ and lim
𝑥↑2
1
(𝑥 − 2)2= +∞ ⇒ lim
𝑥→2
1
(𝑥 − 2)2= ∞: 𝑓(𝑥) has a vertical aymptote 𝑥 = 2
Considering 𝑓(𝑥) =1
𝑥 − 4, then lim
𝑥→4
1
𝑥 − 4 𝐝𝐨𝐞𝐬 𝐧𝐨𝐭 𝐞𝐱𝐢𝐬𝐭 since lim
𝑥↓4
1
𝑥 − 4= +∞ and lim
𝑥↑4
1
𝑥 − 4= −∞
But the graph of 𝑓(𝑥) =1
𝑥−4 does have a vertical asymptote x = 4.
Limits at infinity describe the behaviour of a function if x approaches infinity (∞) or negative infinity (-∞):
clearly, if x attains arbitrarily large values, the value of 1
𝑥 approaches 0:
lim𝑥→∞
1
𝑥= 0 and lim
𝑥→−∞ 1
𝑥= 0 : the function is said to have a horizontal asymptote 𝑦 = 0
Example: 𝑓(𝑥) =4𝑥2−5𝑥+3
𝑥2 has a horizontal asymptote y = 4 since
lim𝑥→∞
4𝑥2−5𝑥+3
𝑥2= lim
𝑥→∞(4 −
5
𝑥+
3
𝑥2) = lim
𝑥→∞4 − lim
𝑥→∞
5
𝑥+ lim𝑥→∞
3
𝑥2 = 4 – 0 + 0 = 4
In this example we used one of the rules for limits below: lim𝑥→𝑎
[𝑓(𝑥) + 𝑔(𝑥)] = lim𝑥→𝑎
𝑓(𝑥) + lim𝑥→𝑎
𝑔(𝑥)
A summary of (informal) definitions and properties of limits:
the graph y=f(x) has a horizontal asymptote y = L if
The Squeeze Theorem
If 𝒇(𝒙) ≤ 𝒈(𝒙) ≤ 𝒉(𝒙) for all values of x near a, then: 𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙) = 𝐥𝐢𝐦 𝒙→𝒂
𝒉(𝒙) = 𝑳 ⇒ 𝐥𝐢𝐦 𝒙→𝒂
𝒈(𝒙) = 𝑳
Example: −𝑥2 ≤ 𝑥2 cos (1
𝑥) ≤ 𝑥2 for x close to 0: lim
𝑥→0 −𝑥2 = lim
𝑥→0𝑥2 = 0 ⇒ lim
𝑥→0𝑥2 cos (
1
𝑥) = 0
Tools for solving limits as 𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙)
1. Direct substitution rule: if 𝑓(𝑥) is a polynomial 𝑃(𝑥) or a rational function 𝑃(𝑥)
𝑄(𝑥) with 𝑄(𝑎) ≠ 0, then
𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙) = 𝒇(𝒂)
2. Use the limit laws above, if possible. Examples combining 1. and 2.: lim𝑥→1
(𝑥3 − 4𝑥2) = lim𝑥→1
𝑥3 − 4 lim𝑥→1
𝑥2 = 1 − 4 × 1 = −3
lim𝑥→1
𝑥3−4𝑥2
𝑥−4=
lim𝑥→1
(𝑥3−4𝑥2)
lim𝑥→1
(𝑥−4)=
1−4
1−4= 1 and lim
𝑥→0
𝑥3−4𝑥2
𝑥−4=
0
−4= 0
3. Factor rational functions: if substitution gives an undetermined value 0
0, factoring can be a solution,
e.g.: 𝑥3−4𝑥2
𝑥−4 gives
0
0 , if x = 4. lim
𝑥→4 𝑥3−4𝑥2
𝑥−4= lim
𝑥→4
𝑥2{𝑥−4}
𝑥−4= lim𝑥→4
𝑥2 =16
4. Apply the “root-trick”: If the limit of a ratio has the form lim𝑥→𝑎
…..
√𝑥−√𝑎 multiply by
√𝑥+√𝑎
√𝑥+√𝑎 and try to
simplify the resulting ratio lim𝑥→𝑎
(…..)(√𝑥+√𝑎)
𝑥−𝑎 by factoring, e.g.:
lim𝑥→9
𝑥2−10𝑥+9
√𝑥−3= lim
𝑥→9 (𝑥2−10𝑥+9)(√𝑥+3)
𝑥−9= lim
𝑥→9 (𝑥−9)(𝑥−1)(√𝑥+3)
𝑥−9= lim
𝑥→9 (𝑥−1)(√𝑥+3)
1=
8×6
1= 48
If 𝑓(𝑥) =𝑥−1
2𝑥−√3+𝑥2 then 𝑓(1) has form
0
0: we can try the root trick:
lim𝑥→1
f(x) = lim𝑥→1
(𝑥−1)
(2𝑥−√3+𝑥2)×(2𝑥+√3+𝑥2 )
(2𝑥+√3+𝑥2 )
= lim𝑥→1
(𝑥−1)(2𝑥+√3+𝑥2)
4𝑥2−(3+𝑥2)= lim
𝑥→1
(𝑥−1)(2𝑥+√3+𝑥2)
3(𝑥−1)(𝑥+1)= lim
𝑥→1
(2𝑥+√3+𝑥2)
3(𝑥+1)=
2+√4
3×2=
2
3
5. Determining the sign (+∞ or -∞), if substitution of x = a in 𝒇(𝒙) leads to the form 𝒄
𝟎 where c ≠ 0:
Compute the limits from the right and from the left of a,
6. Limits at infinity of a rational function 𝑷(𝒙)
𝑸(𝒙): divide by the highest power of x in the denominator,
e.g.: lim𝑥→∞
𝑥3−4𝑥2
3𝑥4−4= lim𝑥→∞
1
𝑥 −
4
𝑥2
3 −4
𝑥4 =
lim𝑥→∞
1
𝑥 − lim
𝑥→∞ 4
𝑥2
lim𝑥→∞
3 − lim𝑥→∞
4
𝑥4 =
0−0
3−0= 0 a horizontal asymptote y = 0
lim𝑥→−∞
𝑥3 − 4𝑥2
3𝑥2 − 4= lim𝑥→−∞
𝑥−4
3 − 4
𝑥2 =
lim𝑥→−∞
𝑥 − lim𝑥→−∞
4
lim𝑥→−∞
3 − lim𝑥→−∞
4
𝑥2 =
−∞−4
3−0= −∞ and lim
𝑥→∞
𝑥3−4𝑥2
3𝑥2−4=∞ (no horiz. as.)
In general: if substitution of x = a or if x approaches ∞ or -∞, leads to
an 𝐢𝐧𝐝𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐚𝐭𝐞 𝐟𝐨𝐫𝐦 𝟎
𝟎 or ±
∞
∞ , then a further investigation or “trick: is necessary.
a form 𝒄
𝟎 where c ≠ 0, then we have a vertical asymptote
The formal definition of 𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙) = 𝑳
We applied the limits, knowing it means that f(x) will be (arbitrarily) near L if x takes on values near a. To give a mathematically precise definition of this informal description, we will consider an interval
(𝐿 − 휀, 𝐿 + 휀) for 𝑓(𝑥) and an interval (𝑎 − 𝛿, 𝑎 + 𝛿) for 𝑥.
𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙) = 𝑳 ⟺ for all 𝜺 > 0, there is a 𝜹 > 0 such that: 0 < |𝒙 − 𝒂| < 𝜹 ⇒ |𝒇(𝒙) − 𝑳| < 𝜺
We solved the tangent problem on page 1 (“Find the slope of the tangent of 𝑦 = 𝑥2 at the point (1, 1)”) by
reasoning that lim𝑥→1
𝑥2−1
𝑥−1= 2 . We will now use the formal definition to proof this is true:
Proof: assume that “x close to 1” means |𝑥 − 1| < 1
Choose an arbitrary number 휀 > 0, then x should be such that |𝒇(𝒙) − 𝑳| < 휀.
|𝑓(𝑥) − 𝐿| = |𝑥2−1
𝑥−1− 2| = |
𝑥2−1−2(𝑥−1)
𝑥−1| = |
𝑥2−2𝑥+1
𝑥−1| = |
(𝑥−1)2
𝑥−1| = |𝑥 − 1| < 휀
So if we choose 𝛿 < 휀 (e. g. 𝛿 =1
2휀) we have proven: |𝑥 − 1| < 𝛿 ⇒ |
𝑥2−1
𝑥−1− 2| = |𝑥 − 1| < 𝛿 < 휀
Infinite limit: 𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙) = ∞ ⟺ for all 𝑹 > 0, there is a 𝜹 > 0 such that: 0 < |𝒙 − 𝒂| < 𝛿 ⇒ 𝑓(𝒙) > 𝑅
Continuity of functions:
Definition of continuity of a function: 𝒇 is continuous at 𝒂 if 𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙) = 𝒇(𝒂) .
This definition implies: 1. 𝑓(𝑥) is defined at 𝑎 , 2. lim𝑥→𝑎
𝑓(𝑥) exists and 3. lim𝑥→𝑎
𝑓(𝑥) 𝑒𝑞𝑢𝑎𝑙𝑠 𝑓(𝑎)
Discontinuity of 𝑓 at 𝑎 is either
1. an infinite discontinuity: the line 𝑥 = 𝑎 is a vertical asymptote or
2. a jump discontinuity: both lim𝑥↓𝑎 𝑓(𝑥) and lim
𝑥↑𝑎 𝑓(𝑥) exist and are not equal 𝐨𝐫
3. a removable discontinuity at a: lim𝑥→𝑎
𝑓(𝑥) exists, but is not equal to 𝑓(𝑎), if 𝑓 is defined at 𝑎
The discontinuity is removed by (re)defining 𝑓(𝑎) = lim𝑥→𝑎
𝑓(𝑥)
(see the example at the end for the graphical meaning of these discontinuities) 𝑓 is continuous from the left if lim
𝑥↑𝑎 𝑓(𝑥) = 𝑓(𝑎) and continuous from the right if lim
𝑥↓𝑎 𝑓(𝑥) = 𝑓(𝑎).
𝑓 is continuous on an interval ⇔ 𝑓 is continuous in every number of the interval. If the interval is closed, e.g. [𝑎, 𝑏], 𝑓 is continuous at every 𝑥 in (𝑎, 𝑏), continuous from the right in 𝑎 and
continuous from the left in 𝑏.
If 𝑓 and 𝑔 are continuous functions at a, then combinations of 𝑓 and 𝑔 are continuous at 𝑎:
𝑐𝑓, 𝑓 + 𝑔, 𝑓– 𝑔, 𝑓 × 𝑔 and, if 𝑔(𝑎 ) ≠ 0, 𝑓/𝑔
Functions that are continuous on their domain: polynomial, rational, root, trigonometric, exponential,
logarithmic functions and all inverse functions of continuous functions.
Continuity of a composite function 𝑓 ∘ 𝑔:
If g is continuous at a and f is continuous at g(a),then 𝑓 ∘ 𝑔 (x) = 𝑓 [𝑔(𝑥)] is continuous at 𝑎.
The Intermediate Value Theorem
If 𝒇 is continuous on [𝒂, 𝒃] and 𝑵 is an arbitrary number between 𝒇(𝒂) and 𝒇(𝒃), then there is a c such that 𝒇(𝒄) = 𝑵.
Some special limits:
At infinity lim𝑥→∞
𝑥𝑛 = ∞, 𝑖𝑓 𝑛 > 0 lim𝑥→∞
1
𝑥𝑛= 0, 𝑖𝑓 𝑛 > 0 lim
𝑥→∞𝑎𝑥 = ∞, if 𝑎 > 1
lim𝑥→−∞
𝑥𝑛 = −∞ for odd n > 0 lim𝑥→∞
𝑎𝑥 = 0 if 0 < a < 1
Trigonometry lim𝑥→0
sin (𝑥)
𝑥= 1 lim
𝑥→0
1 − cos (𝑥)
𝑥= 0 lim
𝑥→±∞𝑡𝑎𝑛−1(𝑥) = ±
1
2𝜋
Derivatives and differentiability
The rate of change of the function 𝒇(𝒙) on the interval (a, x) is ∆𝒚
∆𝒙=
𝒇(𝒙)−𝒇(𝒂)
𝒙−𝒂
Note that if x < a the interval is (x, a) and the rate of change does not alter: 𝒇(𝒂)−𝒇(𝒙)
𝒂−𝒙=
𝒇(𝒙)−𝒇(𝒂)
𝒙−𝒂
The instantaneous rate of change of 𝒚 = 𝒇(𝒙) at 𝒂 is the limit of the rate of change as 𝒙 approaches 𝒂.
Graphically this can be interpreted as the slope of the tangent line of the curve y = f(x) at point (a, f(a))
slope m = 𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙)−𝒇(𝒂)
𝒙−𝒂 (if the limit exists) or, substituting x = a+h: m = 𝐥𝐢𝐦
𝒉→𝟎
𝒇(𝒂+𝒉)−𝒇(𝒂)
𝒉
This slope m depends on the value of a: we will therefore use the notation 𝒇′(𝒂) = 𝐥𝐢𝐦𝒉→𝟎
𝒇(𝒂+𝒉)−𝒇(𝒂)
𝒉
Example: Find the slope of the tangent at (a, f(a)), if 𝑓(𝑥) = 𝑥2, now for arbitrary a (not only for a = 1)
𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ = lim
ℎ→0
(𝑎+ℎ)2−𝑎2
ℎ= lim
ℎ→0
2𝑎ℎ+ℎ2
ℎ= lim
ℎ→0
ℎ(2𝑎+ℎ)
ℎ= lim
ℎ→0 (2𝑎 + ℎ) = 2𝑎
The derivative of 𝑓(𝑥) = 𝑥2 is 𝑓′(𝑥) = 2𝑥: e.g. at x = 1 the slope of the tangent is 𝑓′(1) = 2.
In general: the derivative of the function 𝒚 = 𝒇(𝒙) at a value 𝒙 is: 𝒇′(𝒙) = 𝐥𝐢𝐦𝒉→𝟎
𝒇(𝒙+𝒉)−𝒇(𝒙)
𝒉
(If we use the notations ∆𝑦 = 𝑓(𝑥 + ℎ) − 𝑓(𝑥) and ∆𝑥 = ℎ, we will use the notation 𝑑𝑦
𝑑𝑥= lim
∆𝑥→0 ∆𝑦
∆𝑥 )
Notations for a derivative of function 𝑦 = 𝑓(𝑥): 𝑓′(𝑥) =𝑑𝑦
𝑑𝑥=
𝑑𝑓
𝑑𝑥=
𝑑
𝑑𝑥𝑓(𝑥)
Differentiability
𝒇 is differentiable at 𝒂 if 𝑓 `(𝑎) exists and
𝒇 is differentiable at (𝒂, 𝒃) if f `(x) exists for all 𝑥 ∈ (𝑎, 𝑏): then 𝑓 `(𝑥) can be seen as a function on (𝑎, 𝑏)
Property: 𝑓 is differentiable at a => 𝑓 is continuous at 𝒂 (not vice versa!)
For applications this property means that continuity at 𝒂 is a condition for differentiability at 𝒂.
- If the function is not defined at 𝑎 it cannot be continuous at 𝑎 nor differentiable at 𝑎.
- If the function is continuous at a it could be non differentiable at a,
e.g. if the graph at 𝑦 = 𝑓(𝑎) is not “smooth” but has a sharp corner or has a vertical tangent line:
then the derivative (the limit at 𝑎) does not exist.
All polynomials 𝑃(𝑥) are continuous and differentiable at all real values 𝑎
The second derivative is the derivative of the derivative: 𝑓′′(𝑥) =𝑑