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

𝑑𝑥𝑓′(𝑥) =

𝑑

𝑑𝑥[𝑑𝑓

𝑑𝑥] =

𝑑2𝑓

𝑑𝑥2=

𝑑2𝑦

𝑑𝑥2

Asymptotes:

Exercises Limits, Continuity and derivatives (Stewart Ch. 2) Graph 1 Graph 2 Graph 3

Graph 4:

2.1 The graph of 𝑓 is shown in graph 1. State the value of each quantity if is exists.

If it does not exist, explain why. a. lim𝑥→0

𝑓(𝑥)

b. lim𝑥↑3

𝑓(𝑥) c. lim𝑥↓3

𝑓(𝑥) d. lim𝑥→0

𝑓(𝑥) e. 𝑓(3)

2.2

Like exercise 1, now for graph 2 a. lim𝑥→1

𝑓(𝑥)

b. lim𝑥↓5

𝑓(𝑥) c. lim𝑥↑5

𝑓(𝑥) d. lim𝑥→5

𝑓(𝑥) e. 𝑓(5)

2.3 Like exercise 1, now for graph 3

of the function g(t) a. lim

𝑡↑0𝑔(𝑡) b. lim

𝑡↓0𝑔(𝑡) c. lim

𝑡→0𝑔(𝑡)

d. lim𝑡↑2

𝑔(𝑡) e. lim𝑡↓2

𝑔(𝑥) f. lim𝑡→2

𝑔(𝑥) g. 𝑔(2)

2.4 For graph 4 of the function f state the following a. lim𝑥→6

𝑓(𝑥) b. lim𝑥→0

𝑓(𝑥)

c. lim𝑥↑6

𝑓(𝑥) d. lim𝑥↓6

𝑓(𝑥) e. The equations of the vertical asymptotes

2.5 Sketch the graph of the following function and use it to

determine the values of a for which lim𝑥→𝑎

𝑓(𝑥) exists 𝑓(𝑥) = {

2 − 𝑥 𝑖𝑓 𝑥 < −1 𝑥 𝑖𝑓 − 1 ≤ 𝑥 < 1

(𝑥 − 1)2 𝑖𝑓 𝑥 > 1

2.6

Guess the limit (if it exists) by evaluating the function for x = 2.1, 2.001, 1.9 and

1.99.First check that substituting x = 2 gives the indeterminate form 00.

lim𝑥→2

𝑥2−2𝑥

𝑥2−𝑥−2

2.7 Guess the limit (if it exists) by evaluating the function for x= 0.5, 0.01 and x= -0.5,

-0.01. First check that substituting x = 0 gives you the indeterminate form 00.

lim𝑥→0

𝑒𝑥 − 1 − 𝑥

𝑥2

2.8 Determine lim

𝑥↑1

1

𝑥3 − 1 and lim

𝑥↓1

1

𝑥3 − 1 by

a. Evaluating the function

for values close to 1.

b. Reasoning (function

values if x is close to 1)

2.9 Evaluate the limit and justify each step by indicating the appropriate Limit Law

𝐚. lim𝑥→2

5𝑥2 − 2𝑥 + 3 𝐛. lim

𝑥→−1

𝑥 − 2

𝑥2 + 4𝑥 − 3

𝐜. lim𝑡→−1

(𝑡2 − 1)3(𝑡 + 3)5 𝐝. lim𝑥↑4 √16 − 𝑥2

2.10 a. What is wrong in the

following equation? 𝑥2 + 𝑥 − 6

𝑥 − 2= 𝑥 + 3 𝐛. But why is lim

x→2

𝑥2 + 𝑥 − 6

𝑥 − 2= lim

x→2𝑥 + 3 correct?

2.11 Evaluate the limit if it exists

𝐚. lim𝑥→2

𝑥2 + 𝑥 − 6

𝑥 − 2 𝐛. lim

𝑥→2 𝑥2 − 𝑥 + 6

𝑥 − 2 𝐜. lim

𝑥→4

𝑥2 − 4𝑥

𝑥2 − 3𝑥 − 4 𝐝. lim

ℎ→0 (4 + ℎ)2 − 16

ℎ

𝐞. limℎ→0

(2 + ℎ)3 − 8

ℎ 𝐟. lim

𝑡→9 9 − 𝑡

3 − √𝑡

𝐠. lim𝑥→−4

14+ 1𝑥

4 + 𝑥

𝐡. lim𝑥→9

𝑥2 − 81

√𝑥 − 3

2.12 Use the Squeeze Theorem to

find the limit.

𝐚. Find lim𝑥→4

𝑓(𝑥) , if 4𝑥 − 9 ≤ 𝑓(𝑥) ≤ 𝑥2 − 4𝑥 + 7 (for 𝑥 > 0)

𝐛. Find lim𝑥→1

𝑔(𝑥) , if 2𝑥 ≤ 𝑔(𝑥) ≤ 𝑥4 − 𝑥2 + 2 (for all 𝑥)

2.13 Show that lim

𝑥→0𝑥4𝑐𝑜𝑠 (

2

𝑥) = 0

2.14 Find the limit if it exists.

If not, explain why. 𝐚. lim

𝑥→−6

2𝑥 + 12

|𝑥 + 6| 𝐛. lim

𝑥↑0 (1

𝑥+1

|𝑥|) 𝐜. lim

𝑥↓0 (1

𝑥+1

|𝑥|)

2.15

If 𝑓(𝑥) =

{

𝑥 𝑖𝑓 𝑥 < 1 3 𝑖𝑓 𝑥 = 1 𝑥 − 1 𝑖𝑓 1 < 𝑥 ≤ 2

𝑥 − 3 𝑖𝑓 𝑥 > 2

, find (if it exists)

𝐚. lim𝑥↑1

𝑔(𝑥) 𝐛. lim𝑥→1

𝑔(𝑥)

c. g(1) 𝐝. lim𝑥↓2

𝑔(𝑥)

𝐞. lim𝑥↑2

𝑔(𝑥) 𝐟. lim𝑥→2

𝑔(𝑥)

g. Sketch the graph of g.

2.16 Explain why the function is discontinuous at the given number 𝑥 = 𝑎 and sketch the graph of the function: 𝐚. 𝑓(𝑥) = ln |𝑥 − 2|

if 𝑎 = 2 𝐛. 𝑓(𝑥) = {

1

𝑥 − 1 if 𝑥 ≠ 1

2 if 𝑥 = 1

𝑎 = 1

𝐜. 𝑓(𝑥) = {𝑒𝑥 if 𝑥 < 0𝑥2 if 𝑥 ≥ 0

𝑎 = 0

𝐛. 𝑓(𝑥) = {cos 𝑥 if 𝑥 < 0 0 if 𝑥 = 01 − 𝑥2 if 𝑥 > 0

𝑎 = 0

2.17

State the domain of each function and explain why the function is continuous on this domain.

𝐚. 𝑅(𝑥) = 𝑥2 + √2𝑥 − 1 𝐛. ℎ(𝑥) = 𝑠𝑖𝑛(𝑥)

𝑥 + 1

𝐜. 𝐺(𝑡) = 𝑙𝑛(𝑡4 − 1)

2.18 Find the numbers at which is discontinuous. At which of these numbers is f continuous from the right or left?

𝐚. 𝒇(𝒙) = {1 + 𝑥2 if 𝑥 ≤ 0 2 − 𝑥 if 0 < 𝑥 ≤ 2(𝑥 − 2)2 if 𝑥 > 2

𝐛. 𝑓(𝑥) = {𝑥 + 1 if 𝑥 ≤ 1𝑥2 if 1 < 𝑥 < 3

√𝑥 − 3 if 𝑥 ≥ 3

𝐜. 𝑓(𝑥) = { x + 2 if 𝑥 < 0 𝑒𝑥 if 0 ≤ 𝑥 ≤ 12 − 𝑥 if 𝑥 > 1

2.19 Use the Intermediate Value Theorem to show that there is a root for the given interval (or find such an interval)

𝐚. 𝑥4 + 𝑥 − 3 = 0; (1,2) 𝐛. 𝑐𝑜𝑠(𝑥) = 𝑥; (0,1) 𝐜. 𝑙𝑛(𝑥) = 𝑒−𝑥; (1,2) 𝐝. 𝑒−𝑥 = 2 − 𝑥 2.20

Guess the value of the limit lim𝑥→∞

𝑥2

2𝑥 by evaluating it for 𝑥 = 1, 10, 20 and 100

2.21 Evaluate the limit and justify each step by indicating the

appropriate property of limits that is applied. lim𝑥→∞

3𝑥2 − 𝑥 + 6

2𝑥2 + 5𝑥 − 8

2.22

Find the limit

𝐚. lim𝑥→∞

1

2𝑥 + 3 𝐛. lim

𝑥→∞ 3𝑥 + 5

𝑥 − 4 𝐜. lim

𝑡→−∞

6𝑡2 + 5𝑡

(1 − 𝑡)(2𝑡 − 3)

𝐝. lim𝑥→∞

𝑥3 + 5𝑥

2𝑥3 − 𝑥2 + 4 𝐞. lim

𝑥→∞ √9𝑥6 − 𝑥

𝑥3 + 1

𝐟. lim𝑥→∞

𝑐𝑜𝑠 𝑥 𝐠. lim𝑥→∞

√9𝑥2 + 𝑥 − 3𝑥

𝐡. lim𝑥→∞

𝑥 + 𝑥3 + 𝑥5

1 − 𝑥2 + 𝑥4

𝐢. lim𝑥→∞

𝑥4 + 𝑥5 𝐣. lim𝑥→∞

1 − 𝑒𝑥

1 + 2𝑒𝑥

𝐤. lim𝑥→∞

𝑒−2𝑥𝑐𝑜𝑠 𝑥

2.23 Find the horizontal and

vertical asymptotes 𝐚. 𝑦 =1 − 𝑒𝑥

1 + 2𝑒𝑥 𝐛. 𝑦 =

1 + 𝑥4

𝑥2 − 𝑥4 𝐜. 𝑦 =

𝑥3 − 𝑥

𝑥2 − 6𝑥 + 5

2.24 Find lim𝑥→∞

𝑓(𝑥), if for all 𝑥 > 1: 10𝑒𝑥−21

2𝑒𝑥< 𝑓(𝑥) <

5√𝑥

√𝑥−1

2.25 Find the derivative and an equation of the tangent line to

the curve at the given point 𝐚. 𝑦 = √𝑥 ; (1,1) 𝐛. 𝑦 =

𝑥 − 1

𝑥 − 2; (3,2)

2.26 If a ball is thrown into the air with a velocity of 10 m/s its height in meters after 𝑡 seconds is given by:

𝒚 = 𝟏𝟎𝒕 – 𝟒. 𝟗𝒕𝟐. Find the velocity (the derivative) when 𝑡 = 2.

2.27 Find 𝑓 `(𝑎) 𝐚. 𝑓(𝑥) = 3 − 2𝑥 + 4𝑥2 𝐛. 𝑓(𝑡) =

2𝑡 + 1

𝑡 + 3

2.28 A particle moves along a straight with the equation y = 1

𝑥 – x (𝑦 is in m and x in s). Find the velocity at x = 5

2.29 Match the graph of each of the functions (a)-(d) with the

graph of its derivatives given in I-IV.

2.30 Find the derivative of the

function using the definition

of derivative. State the

domain of the function and

the domain of its derivative

𝐚. 𝑓(𝑥) = 1

2𝑥 −

1

3

𝐛. 𝑓(𝑡) = 5𝑡 − 9𝑡2

𝐜. 𝑓(𝑥) = 𝑥 + √𝑥

2.31 True or false quiz:

𝐚. lim𝑥→4

(2𝑥

𝑥 − 4−

8

𝑥 − 4) = lim

𝑥→4 2𝑥

𝑥 − 4 − lim

𝑥→4 8

𝑥 − 4

𝐛. lim𝑥→1

𝑥2 + 6𝑥 − 7

𝑥2 + 5𝑥 − 6 =

lim𝑥→1

𝑥2 + 6𝑥 − 7

lim𝑥→1

𝑥2 + 5𝑥 − 6

𝐜. lim𝑥→1

𝑥 − 3

𝑥2 + 2𝑥 − 4 =

lim𝑥→1

𝑥 − 3

lim𝑥→1

𝑥2 + 2𝑥 – 4

d. if lim𝑥→5

𝑓(𝑥) = 2 and lim𝑥→5

𝑔(𝑥) = 0 ,

then lim𝑥→5

[𝑓(𝑥)/𝑔(𝑥)] does not exist.

e. if lim𝑥→5

𝑓(𝑥) = 0 and lim𝑥→5

𝑔(𝑥) = 0 ,

then lim𝑥→5

[𝑓(𝑥)/𝑔(𝑥)] does not exist.

f. If lim𝑥→6

[𝑓(𝑥) 𝑔(𝑥)] exists, then the limit must be f(6)g(6).

g. If 𝑝 is a polynomial then lim𝑥→𝑏

𝑝(𝑥) = 𝑝(𝑏)

h. If lim𝑥→0

𝑓(𝑥) = ∞ and lim𝑥→0

𝑔(𝑥) = ∞ ,

then lim𝑥→0

[𝑓(𝑥) −𝑔(𝑥)] = 0

i. A function can have two different horizontal asymptotes.

j. If 𝑥 = 1 is a vertical asymptote of 𝑦 = 𝑓 (𝑥), then 𝑓(1) is not defined.

k. If 𝑓(1) > 0 and 𝑓(3) < 0,

there exists a 𝑐 such that 1 < 𝑐 < 3 and 𝑓(𝑐) = 0.

l. If 𝑓 is continuous at 𝑎, then f is differentiable at 𝑎.

Answers Exercises Limits, Continuity and derivatives (Stewart Ch. 2)

2.1 a. 3 b. 4 c. 2 d. 3 e. 3

2.2 a. – (does not exist) b. 4 c. 4 d. 4 e. - (not defined)

2.3 a. -1 b. -2 c. – (d.n.e.) d. 2 e. 0 f. – (d.n.e) g. 1

2.4 a. d.n.e. b. ∞ c. -∞ d. ∞ e. 𝑥 = −7, 𝑥 = −3, 𝑥 = 0, 𝑥 = 6

2.5 𝐥𝐢𝐦𝒙→𝒂

𝒇(𝒙) exists for all 𝑎 ≠ ±1

2.6 0.677419, 0,666778, 0.655172, 0.666566: our guess is that the limit equals 2/3

2.7 0.594885, 0.501671, 0.426123, 0.498337: our guess is that the limit equals ½

2.8 a. e.g. 𝑓(0.99) = −33.7 and 𝑓(1.01) = 33.0 b. if 𝑥 is slightly smaller than 1, then 𝑥3 − 1 will be

a negative number close to zero: its reciprocal is large negative. So 𝐥𝐢𝐦𝒙↑𝟏

𝟏

𝒙𝟑−𝟏 = −∞

2.9 Use the a. sum law b. quotient and sum law resp. c. product and power law d. root law.

2.10 a. (x – 2) can only be cancelled if 𝑥 ≠ 2 b. the limit as 𝑥 → 2 means 𝑥 ≠ 2!

2.11 a. 5 b. does not exist c. 4/5 d. 8 e. 12 f. 6 g. -1/16 h. 108

2.12 a. 7 b. 2

2.13 Use the squeeze theorem and -1 ≤ cos (y) ≤ 1

2.14 a. No, limit from the left (-2) and from the right (2) are not equal b. No (∞) c. Yes, limit = 0

2.15 a. 1 b. 1 c. 3 d. -2 e. -1 f. does not exist (limits from the left and the right are not equal)

2.16 a. asymptotic discontinuity b. asymptotic disc. c. jump disc. d. lim𝑥↑0

𝑓(𝑥) = lim𝑥↓0

𝑓(𝑥) ¹ 𝑓(0)

2.17 a. Polynomials and root functions are continuous on their domain and so is the sum of the two

b. Trigonometric functions and polynomials are continuous on their domain and so is the quotient.

c. logarithm and polynomials are continuous on their domain and so is the composition of them.

2.18 a. discontinuous at 𝑥 = 0 (but continuous from the left)

b. discontinuous at 𝑥 = 0 and 𝑥 = 3, but continuous from the left at 𝑥 = 0 and from the right at 𝑥 = 3

c. discontinuous at 𝑥 = 0 and 𝑥 = 1, but continuous from the left at 1 and from the right at 0

2.19 a. 𝑓(1) = −1 and 𝑓(2) = 15 b. 𝑓(𝑥) = cos(x) – x: 𝑓(0) = 1 and 𝑓(1) = 𝑐𝑜𝑠(1) − 1 < 0

c. 𝑓(𝑥) = 𝑙𝑛(𝑥)– 𝑒𝑥 : 𝑓(1) =– 𝑒−1 < 0 and 𝑓(𝑥) = 𝑙𝑛(2) – 𝑒2 > 0

d. 𝑓(𝑥) = 𝑒𝑥– 2 + 𝑥: 𝑓(0) = −1 and 𝑓(3) = 𝑒3 + 1 > 0

2.20 𝑓(𝑥) = 𝑥2/2𝑥:𝑓(1) = 0.5, 𝑓(10) = 0.0977,𝑓(20) = 0.000381,𝑓(100) = 7.8886´10−27: limit ≈ 0

2.21 Start by dividing numerator and denominator by 𝑥2: apply quotient and addition laws of limits:

lim𝑥→∞

3𝑥2 − 𝑥 + 6

2𝑥2 + 5𝑥 − 8= lim

𝑥→∞ 3 −

1𝑥+6𝑥2

2 +5𝑥 −

8𝑥2

=lim𝑥→∞

[3 −1𝑥 +

6𝑥2 ]

lim𝑥→∞

[2 +5𝑥 −

8𝑥2]=

lim𝑥→∞

3 − lim𝑥→∞

1𝑥 + lim

𝑥→∞ 6𝑥2

lim𝑥→∞

2 + lim𝑥→∞

5𝑥 − lim

𝑥→∞ 8𝑥2

= 3

2

2.22 a. 0 b. 3 c. -3 d. ½ e. 3 f. does not exist g. 1/6 h. ∞ i. -∞ use 𝒙𝟒 + 𝒙𝟓 = 𝒙𝟓 (𝟏

𝒙+ 𝟏) j. -½ k. 0: squeeze

2.23 a. y = 1, x = -4 b. y = -1, x = -1, x = 0, x = 1 c. x = 5 (x = 1 is a removable discontinuity)

2.24 5 (apply the squeeze theorem)

2.25 a. 𝑚 = 𝑓 `(1) = ½: 𝑦 = ½ 𝑥 +½ b. 𝑚 = 𝑓 `(3) = −1: 𝑦 = −𝑥 + 5

2.26 The velocity (the derivative of the position function) at 𝑡 = 2 is – 9.6

2.27 a. −2 + 8𝑎 b. 5/(𝑡 + 3)2

2.28 The derivative at 𝑥 = 5 is -26/25 (the velocity is 26/25 m/s)

2.29 (a) → II , (b) → IV, (c) → I, (d) → III

2.30 a. 𝑓 `(𝑥) = ½, 𝐷𝑓 = 𝐷𝑓` = ℝ

b. 𝑓 `(𝑡) = 15 − 18𝑡, 𝐷𝑓 = 𝐷𝑓` = ℝ

c. 𝑓 `(𝑥) = 1 + 1

2√𝑥 , 𝐷𝑓 = [0,∞) and 𝐷𝑓` = (0,∞)

2.31 a. False b. False c. True d. True e. False f. False

g. True h. False i. True j. False k. False l. False