Practice Final #4 (focus on short answer)

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McGill University, MATH 139

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Short Answer Questions

Short Answer
A1
Difficulty: 2/10
Evaluate the following limit:
lim⁡h→0(−4+h)2−16h \lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h}

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limits: general

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Question

Evaluate the following limit:
lim⁡h→0(−4+h)2−16h \lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h}

Step 1: Expand the numerator


(−4+h)2−16=((−4)2+2(−4)(h)+h2)−16=16+2(−4)(h)+h2−16=2(−4)(h)+h2=−8h+h2 \begin{align*}
&(-4 + h)^2 - 16\\
&= ((-4)^2 + 2(-4)(h) + h^2) - 16 \\
&= 16 + 2(-4)(h) + h^2 - 16 \\
&= 2(-4)(h) + h^2 \\
&= -8h + h^2
\end{align*}


Step 2: Substitute into the original limit


lim⁡h→0(−4+h)2−16h=lim⁡h→0−8h+h2h=lim⁡h→0h(−8+h)h=lim⁡h→0(−8+h) \begin{align*}
& \lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h} \\
&= \lim_{h \to 0} \frac{-8h + h^2}{h} \\
&= \lim_{h \to 0} \frac{h(-8 + h)}{h} \\
&= \lim_{h \to 0} (-8 + h)
\end{align*}


Step 3: Evaluate the limit

lim⁡h→0(−8+h)=−8+0=−8 \begin{align*}
\lim_{h \to 0} (-8 + h) &= -8 + 0 \\
&= -8
\end{align*}

Therefore:
lim⁡h→0(−4+h)2−16h=−8\boxed{\lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h} = -8}
lim⁡h→0(−4+h)2−16h=−8\boxed{\lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h} = -8}
A2
Difficulty: 1/10
Solve lim⁡t→−3t2−93t2+11t+6. \lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6}.

Exercise Tags

limits: general
Factoring: basic

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Step 1: Factor the expressions


lim⁡t→−3t2−93t2+11t+6 \lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6}

For the numerator:
t2−9=(t−3)(t+3) t^2 - 9 = (t-3)(t+3)

For the denominator:

3t2+11t+6=3t2+9t+2t+6=3t(t+3)+2(t+3)=(3t+2)(t+3) \begin{align*}
3t^2 + 11t + 6 &= 3t^2 + 9t + 2t + 6 \\
&= 3t(t+3) + 2(t+3) \\
&= (3t+2)(t+3)
\end{align*}



Step 2: Cancel common factors


lim⁡t→−3t2−93t2+11t+6 \lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6}
=lim⁡t→−3(t−3)(t+3)(3t+2)(t+3) = \lim_{t \to -3} \frac{(t-3)(t+3)}{(3t+2)(t+3)}
=lim⁡t→−3(t−3)(t+3)(3t+2)(t+3) = \lim_{t \to -3} \frac{(t-3)\cancel{(t+3)}}{(3t+2)\cancel{(t+3)}}
=lim⁡t→−3t−33t+2 = \lim_{t \to -3} \frac{t-3}{3t+2}

Step 3: Evaluate the limit


lim⁡t→−3t−33t+2 \lim_{t \to -3} \frac{t-3}{3t+2}
=−3−33(−3)+2 = \frac{-3-3}{3(-3)+2}
=−6−9+2 = \frac{-6}{-9+2}
=−6−7 = \frac{-6}{-7}
=67 = \frac{6}{7}

Therefore:
lim⁡t→−3t2−93t2+11t+6=67\boxed{\lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6} = \frac{6}{7}}
lim⁡t→−3t2−93t2+11t+6=67\boxed{\lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6} = \frac{6}{7}}
A3
Difficulty: 3/10
Find

lim⁡x→2(tan⁡((x3−x2−x−2)πx2+17x−38))
\lim_{x \to 2} \left( \tan \left( \frac{(x^3 - x^2 - x - 2)\pi }{x^2 + 17x - 38} \right) \right)

Exercise Tags

factoring: grouping
unit circle values
limits: general

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Step 1: Simplify the expression inside the tangent function


To evaluate this limit, start by simplifying the expression within the tangent function. We’ll try factoring the numerator and the denominator.
The numerator x3−x2−x−2 x^3 - x^2 - x - 2 can be factored by grouping terms.

Factoring by Grouping


To factor x3−x2−x−2 x^3 - x^2 - x - 2 , we group the terms in pairs:

x3−x2−x−2=(x3−x2)−(x+2)
x^3 - x^2 - x - 2 = (x^3 - x^2) - (x + 2)


Now, factor out x2 x^2 from the first group and factor out −1 -1 from the second group:

=x2(x−1)−1(x−1)
= x^2(x - 1) - 1(x - 1)


Now we have a common factor of (x−2) (x - 2) :

=(x−2)(x2+x+1)
= (x - 2)(x^2 + x + 1)


The denominator x2+17x−38 x^2 + 17x - 38 factors as:

x2+17x−38=(x−2)(x+19)
x^2 + 17x - 38 = (x - 2)(x + 19)


So, the expression becomes:

tan⁡((x−2)(x2+x+1)(x−2)(x+19)π)
\tan \left( \frac{(x - 2)(x^2 + x + 1)}{(x - 2)(x + 19)} \pi \right)


Step 2: Cancel common terms


Now, cancel the common factor (x−2) (x - 2) from both the numerator and the denominator:

tan⁡(x2+x+1x+19π)
\tan \left( \frac{x^2 + x + 1}{x + 19} \pi \right)


Step 3: Substitute the limit


Now that we’ve simplified the expression, we can substitute x=2 x = 2 directly:

tan⁡(22+2+12+19π)=
\tan \left( \frac{2^2 + 2 + 1}{2 + 19} \pi \right) =


tan⁡(4+2+121π)=
\tan \left( \frac{4 + 2 + 1}{21} \pi \right) =


tan⁡(721π)=
\tan \left( \frac{7}{21} \pi \right) =


tan⁡(π3)
\tan \left( \frac{\pi}{3} \right)


Tip: You need to know the unit circle and the values of common trigonometric functions at special angles, such as π3 \frac{\pi}{3} , to solve problems like this.


Since tan⁡(π3)=3 \tan \left( \frac{\pi}{3} \right) = \sqrt{3} , we have:

3
\boxed{\sqrt{3}}


Prof's perspective


This question is designed to test your ability to handle limits involving trigonometric functions, especially by simplifying rational expressions through factoring. The professor likely included this question to emphasize the importance of **factoring and cancelling terms** before evaluating limits, a common approach in calculus to avoid indeterminate forms.
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3
\boxed{\sqrt{3}}
A4
Difficulty: 3/10
Find the derivative of

sin⁡(e3x+17ln⁡(x2+1))
\sin \left( e^{3x} + 17 \ln(x^2 + 1) \right)

Exercise Tags

Differentiation: general
differentiation: logarithmic

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Step 1: Recognize the need for the chain rule


To differentiate sin⁡(e3x+17ln⁡(x2+1)) \sin \left( e^{3x} + 17 \ln(x^2 + 1) \right) , we’ll apply the chain rule. The chain rule states:

Chain Rule: If f(x)=g(h(x)) f(x) = g(h(x)) , then f′(x)=g′(h(x))⋅h′(x) f'(x) = g'(h(x)) \cdot h'(x)

In this case:
- g(x)=sin⁡(x) g(x) = \sin(x)
- h(x)=e3x+17ln⁡(x2+1) h(x) = e^{3x} + 17 \ln(x^2 + 1)

Step 2: Differentiate the outer function


First, differentiate the outer function g(x)=sin⁡(x) g(x) = \sin(x) :

g′(h(x))=cos⁡(e3x+17ln⁡(x2+1))
g'(h(x)) = \cos \left( e^{3x} + 17 \ln(x^2 + 1) \right)


Step 3: Differentiate the inner function


Now, we differentiate the inner function h(x)=e3x+17ln⁡(x2+1) h(x) = e^{3x} + 17 \ln(x^2 + 1) :

**Derivative of** e3x e^{3x} :

ddx(e3x)=3e3x
\frac{d}{dx} \left( e^{3x} \right) = 3e^{3x}


**Derivative of** 17ln⁡(x2+1) 17 \ln(x^2 + 1) :

Using the chain rule again on ln⁡(x2+1) \ln(x^2 + 1) :

ddx(17ln⁡(x2+1))=17⋅1x2+1⋅2x=34xx2+1
\frac{d}{dx} \left( 17 \ln(x^2 + 1) \right) = 17 \cdot \frac{1}{x^2 + 1} \cdot 2x = \frac{34x}{x^2 + 1}


Combining these, we have:

h′(x)=3e3x+34xx2+1
h'(x) = 3e^{3x} + \frac{34x}{x^2 + 1}


Step 4: Apply the chain rule


Now, we can combine g′(h(x)) g'(h(x)) and h′(x) h'(x) to find the derivative:


ddx(sin⁡(e3x+17ln⁡(x2+1)))=
\frac{d}{dx} \left( \sin \left( e^{3x} + 17 \ln(x^2 + 1) \right) \right) =


cos⁡(e3x+17ln⁡(x2+1))⋅(3e3x+34xx2+1)
\cos \left( e^{3x} + 17 \ln(x^2 + 1) \right) \cdot
\left( 3e^{3x} + \frac{34x}{x^2 + 1} \right)


cos⁡(e3x+17ln⁡(x2+1))⋅(3e3x+34xx2+1)
\boxed{\cos \left( e^{3x} + 17 \ln(x^2 + 1) \right) \cdot \left( 3e^{3x} + \frac{34x}{x^2 + 1} \right)}





Prof's perspective


This question checks your ability to apply the chain rule in complex scenarios involving trigonometric, exponential, and logarithmic functions. The professor likely chose this question to ensure you can differentiate nested functions accurately, a skill that is crucial for handling advanced calculus problems.
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cos⁡(e3x+17ln⁡(x2+1))⋅(3e3x+34xx2+1)
\boxed{\cos \left( e^{3x} + 17 \ln(x^2 + 1) \right) \cdot \left( 3e^{3x} + \frac{34x}{x^2 + 1} \right)}


A5
Difficulty: 3/10
Find ddx(1+exx2+9). \frac{d}{dx} \left( \frac{1 + e^x}{x^2 + 9} \right).

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quotient rule
Differentiation: general

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Step 1: Recognize the need for the quotient rule


To differentiate 1+exx2+9 \frac{1 + e^x}{x^2 + 9} , we use the quotient rule, which states:

Quotient Rule: If f(x)=uv f(x) = \frac{u}{v} , then f′(x)=u′v−uv′v2 f'(x) = \frac{u'v - uv'}{v^2}

In this case:
- u=1+ex u = 1 + e^x
- v=x2+9 v = x^2 + 9

Step 2: Differentiate the numerator and denominator


**Derivative of** u=1+ex u = 1 + e^x :

u′=ddx(1+ex)=ex
u' = \frac{d}{dx} (1 + e^x) = e^x


**Derivative of** v=x2+9 v = x^2 + 9 :

v′=ddx(x2+9)=2x
v' = \frac{d}{dx} (x^2 + 9) = 2x


Step 3: Apply the quotient rule


Now, substitute u u , u′ u' , v v , and v′ v' into the quotient rule formula:



ddx(1+exx2+9)=(ex)(x2+9)−(1+ex)(2x)(x2+9)2
\frac{d}{dx} \left( \frac{1 + e^x}{x^2 + 9} \right) = \frac{(e^x)(x^2 + 9) - (1 + e^x)(2x)}{(x^2 + 9)^2}




Step 4: Simplify the expression (if possible)


Distribute ex e^x and 2x 2x :

=ex⋅x2+9ex−2x−2xex(x2+9)2
= \frac{e^x \cdot x^2 + 9e^x - 2x - 2x e^x}{(x^2 + 9)^2}


ex⋅x2+9ex−2x−2xex(x2+9)2
\boxed{\frac{e^x \cdot x^2 + 9e^x - 2x - 2x e^x}{(x^2 + 9)^2}}
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ex⋅x2+9ex−2x−2xex(x2+9)2
\boxed{\frac{e^x \cdot x^2 + 9e^x - 2x - 2x e^x}{(x^2 + 9)^2}}
A6
Difficulty: 2/10
Find ddx(log⁡3(x)+4x−x2+17). \frac{d}{dx} \left( \log_3(x) + 4^x - x^2 + 17 \right).

Exercise Tags

differentiation: logarithmic
Differentiation: general

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Step 1: Differentiate each term separately


To find the derivative, apply the rules of differentiation to each term individually. We’ll use the following key rules:

Derivative of log⁡a(x) \log_a(x) : 1xln⁡(a) \frac{1}{x \ln(a)} and Derivative of ax a^x : axln⁡(a) a^x \ln(a)

**Derivative of** log⁡3(x) \log_3(x) :

ddx(log⁡3(x))=1xln⁡(3)
\frac{d}{dx} \left( \log_3(x) \right) = \frac{1}{x \ln(3)}


**Derivative of** 4x 4^x :

ddx(4x)=4xln⁡(4)
\frac{d}{dx} \left( 4^x \right) = 4^x \ln(4)


**Derivative of** −x2 -x^2 :

ddx(−x2)=−2x
\frac{d}{dx} \left( -x^2 \right) = -2x


**Derivative of** 17 17 (a constant):

ddx(17)=0
\frac{d}{dx} \left( 17 \right) = 0


Step 2: Combine the results


Now, we can combine each of these derivatives to find the overall derivative:


ddx(log⁡3(x)+4x−x2+17)=1xln⁡(3)+4xln⁡(4)−2x
\begin{align*}
\frac{d}{dx} \left( \log_3(x) + 4^x - x^2 + 17 \right)
&= \frac{1}{x \ln(3)} \\
&\quad + 4^x \ln(4) - 2x
\end{align*}




1xln⁡(3)+4xln⁡(4)−2x\boxed{\frac{1}{x \ln(3)} + 4^x \ln(4) - 2x}
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1xln⁡(3)+4xln⁡(4)−2x\boxed{\frac{1}{x \ln(3)} + 4^x \ln(4) - 2x}
A7
Difficulty: 2/10
Find lim⁡x→5x2−8x+15x2−25. \lim_{x \to 5} \frac{x^2 - 8x + 15}{x^2 - 25}.

Exercise Tags

limits: general
parabolas

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Step 1: Simplify the expression


To evaluate this limit, start by factoring both the numerator and the denominator. Factoring can often help us cancel out terms, especially if direct substitution initially results in an indeterminate form like 00\frac{0}{0}.

The numerator x2−8x+15 x^2 - 8x + 15 factors as:

x2−8x+15=(x−5)(x−3)
x^2 - 8x + 15 = (x - 5)(x - 3)



The denominator x2−25 x^2 - 25 is a difference of squares, which factors as:

x2−25=(x−5)(x+5)
x^2 - 25 = (x - 5)(x + 5)


So, the expression becomes:

(x−5)(x−3)(x−5)(x+5)
\frac{(x - 5)(x - 3)}{(x - 5)(x + 5)}


Step 2: Cancel common terms


Now, cancel the common factor (x−5) (x - 5) from both the numerator and the denominator:

(x−3)(x+5)
\frac{(x - 3)}{(x + 5)}


Step 3: Substitute the limit


Now that we’ve simplified the expression, we can substitute x=5 x = 5 directly:

5−35+5=210=15
\frac{5 - 3}{5 + 5} = \frac{2}{10} = \frac{1}{5}


Key Tip: When faced with indeterminate forms, factor and cancel common terms to simplify the expression.

15
\boxed{\frac{1}{5}}
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15
\boxed{\frac{1}{5}}
A8
Difficulty: 4/10
Compute the following limit:
lim⁡x→π2sec⁡(x)−tan⁡(x)
\lim_{x \to \frac{\pi}{2}} \sec(x) - \tan(x)

Exercise Tags

L'hopitals rule
limits: general
trig identities

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Step 1: Identify the form of the limit


We are tasked with solving:

lim⁡x→π2sec⁡(x)−tan⁡(x)
\lim_{x \to \frac{\pi}{2}} \sec(x) - \tan(x)


As x x approaches π2 \frac{\pi}{2} , both sec⁡(x) \sec(x) and tan⁡(x) \tan(x) tend to infinity, creating the indeterminate form ∞−∞ \infty - \infty . To properly evaluate this limit, we need to manipulate the expression into a form suitable for limit evaluation.

When dealing with indeterminate forms like ∞−∞ \infty - \infty , it’s helpful to simplify the terms into a single fraction.

Step 2: Rewrite the expression in terms of sin⁡(x) \sin(x) and cos⁡(x) \cos(x)


We express sec⁡(x) \sec(x) and tan⁡(x) \tan(x) using sine and cosine functions:

sec⁡(x)=1cos⁡(x),tan⁡(x)=sin⁡(x)cos⁡(x).
\sec(x) = \frac{1}{\cos(x)}, \quad \tan(x) = \frac{\sin(x)}{\cos(x)}.


Now, the limit becomes:

lim⁡x→π2(1cos⁡(x)−sin⁡(x)cos⁡(x)).
\lim_{x \to \frac{\pi}{2}} \left( \frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)} \right).


We can factor out 1cos⁡(x) \frac{1}{\cos(x)} from both terms:

lim⁡x→π21−sin⁡(x)cos⁡(x).
\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin(x)}{\cos(x)}.


Step 3: Analyze the behavior as x→π2 x \to \frac{\pi}{2}


As x x approaches π2 \frac{\pi}{2} :

- sin⁡(π2)=1 \sin\left( \frac{\pi}{2} \right) = 1 .
- cos⁡(π2)=0 \cos\left( \frac{\pi}{2} \right) = 0 .

Thus, the expression approaches the indeterminate form 00 \frac{0}{0} , making it suitable for applying L'Hopital's Rule.

Step 4: Apply L'Hopital's Rule


We differentiate the numerator and the denominator separately:

1. **Differentiate the numerator** 1−sin⁡(x) 1 - \sin(x) :
- The derivative of 1 1 is 0, and the derivative of sin⁡(x) \sin(x) is cos⁡(x) \cos(x) . Thus:

ddx(1−sin⁡(x))=−cos⁡(x).
\frac{d}{dx} \left( 1 - \sin(x) \right) = -\cos(x).


2. **Differentiate the denominator** cos⁡(x) \cos(x) :
- The derivative of cos⁡(x) \cos(x) is −sin⁡(x) -\sin(x) . Thus:

ddx(cos⁡(x))=−sin⁡(x).
\frac{d}{dx} \left( \cos(x) \right) = -\sin(x).


Step 5: Simplify and evaluate the limit


Now, the limit becomes:

lim⁡x→π2−cos⁡(x)−sin⁡(x)=lim⁡x→π2cos⁡(x)sin⁡(x).
\lim_{x \to \frac{\pi}{2}} \frac{-\cos(x)}{-\sin(x)} = \lim_{x \to \frac{\pi}{2}} \frac{\cos(x)}{\sin(x)}.


As x→π2 x \to \frac{\pi}{2} :

- cos⁡(π2)=0 \cos\left( \frac{\pi}{2} \right) = 0 .
- sin⁡(π2)=1 \sin\left( \frac{\pi}{2} \right) = 1 .

Thus, the limit simplifies to:

01=0.
\frac{0}{1} = 0.


Step 6: Final Answer


The value of the limit is:

0.
\boxed{0}.


Prof's perspective


This problem is designed to test your understanding of **trigonometric identities** and how to handle **indeterminate forms** using L'Hopital's Rule.
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The value of the limit is:

0
\boxed{0}
A9
Difficulty: 7/10
Compute the following limit:
lim⁡x→0sinh⁡(x)−xsin⁡(x)−x
\lim_{x \to 0} \frac{\sinh(x) - x}{\sin(x) - x}

Exercise Tags

L'hopitals rule
limits: general
hyperbolic trig functions

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Step 1: Identify the form of the limit


We are tasked with solving:

lim⁡x→0sinh⁡(x)−xsin⁡(x)−x
\lim_{x \to 0} \frac{\sinh(x) - x}{\sin(x) - x}


As x→0 x \to 0 , both sinh⁡(x)−x \sinh(x) - x and sin⁡(x)−x \sin(x) - x approach 0. This results in the indeterminate form 00 \frac{0}{0} , making this a case for applying L'Hopital's Rule.

Remember, L'Hopital's Rule applies when the limit results in either 00 \frac{0}{0} or ∞∞ \frac{\infty}{\infty} .

Step 2: Differentiate the numerator and denominator


We apply L'Hopital's Rule by differentiating the numerator and denominator separately.

1. **Differentiate the numerator** sinh⁡(x)−x \sinh(x) - x :
- The derivative of sinh⁡(x) \sinh(x) is cosh⁡(x) \cosh(x) , and the derivative of x x is 1. Thus:

ddx(sinh⁡(x)−x)=cosh⁡(x)−1
\frac{d}{dx} \left( \sinh(x) - x \right) = \cosh(x) - 1


2. **Differentiate the denominator** sin⁡(x)−x \sin(x) - x :
- The derivative of sin⁡(x) \sin(x) is cos⁡(x) \cos(x) , and the derivative of x x is 1. Thus:

ddx(sin⁡(x)−x)=cos⁡(x)−1
\frac{d}{dx} \left( \sin(x) - x \right) = \cos(x) - 1


Step 3: Apply the limit again


Now, the limit becomes:

lim⁡x→0cosh⁡(x)−1cos⁡(x)−1
\lim_{x \to 0} \frac{\cosh(x) - 1}{\cos(x) - 1}


Evaluating both the numerator and the denominator at x=0 x = 0 :

- cosh⁡(0)=1 \cosh(0) = 1 , so cosh⁡(x)−1→0 \cosh(x) - 1 \to 0 .
- cos⁡(0)=1 \cos(0) = 1 , so cos⁡(x)−1→0 \cos(x) - 1 \to 0 .

We encounter the indeterminate form 00 \frac{0}{0} again, so we apply L'Hopital's Rule a second time.

Applying L'Hopital's Rule multiple times is perfectly valid when necessary.

Step 4: Differentiate again


1. **Differentiate the numerator** cosh⁡(x)−1 \cosh(x) - 1 :
- The derivative of cosh⁡(x) \cosh(x) is sinh⁡(x) \sinh(x) .

ddx(cosh⁡(x)−1)=sinh⁡(x)
\frac{d}{dx} \left( \cosh(x) - 1 \right) = \sinh(x)


2. **Differentiate the denominator** cos⁡(x)−1 \cos(x) - 1 :
- The derivative of cos⁡(x) \cos(x) is −sin⁡(x) -\sin(x) .

ddx(cos⁡(x)−1)=−sin⁡(x)
\frac{d}{dx} \left( \cos(x) - 1 \right) = -\sin(x)


Step 5: Evaluate the limit


Now, the limit becomes:

lim⁡x→0sinh⁡(x)−sin⁡(x)
\lim_{x \to 0} \frac{\sinh(x)}{-\sin(x)}


For small x x , we use the small-angle approximations sinh⁡(x)≈x \sinh(x) \approx x and sin⁡(x)≈x \sin(x) \approx x . Thus:

lim⁡x→0x−x=−1
\lim_{x \to 0} \frac{x}{-x} = -1



For very small values of x x , sin⁡(x) \sin(x) and sinh⁡(x) \sinh(x) behave similarly, leading to:
lim⁡x→0sinh⁡(x)sin⁡(x)=1
\lim_{x \to 0} \frac{\sinh(x)}{\sin(x)} = 1



Step 6: Final Answer


Thus, the value of the limit is:

−1
\boxed{-1}


Prof's perspective


This problem tests your ability to use L'Hopital's Rule effectively and repeatedly. It emphasizes the importance of knowing small-angle approximations for functions like sin⁡(x) \sin(x) and sinh⁡(x) \sinh(x) . Additionally, the exercise aims to develop an intuition for recognizing when two functions behave similarly for small inputs.
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The value of the limit is:

−1
\boxed{-1}
A10
Difficulty: 4/10
Compute the following limit:
lim⁡x→05(1/x)2(1/x2)
\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}}

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limits: with logarithms

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Step 1: Understand the expression


We are tasked with finding the limit:

lim⁡x→05(1/x)2(1/x2)
\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}}


This expression involves two exponential terms: 5(1/x) 5^{(1/x)} and 2(1/x2) 2^{(1/x^2)} .

Step 2: Investigate the behavior as x→0 x \to 0


Let’s analyze the behavior of both terms:

1. The term 5(1/x) 5^{(1/x)} : As x→0 x \to 0 , the exponent 1/x 1/x grows large, making 5(1/x) 5^{(1/x)} very large.

2. The term 2(1/x2) 2^{(1/x^2)} : As x→0 x \to 0 , the exponent 1/x2 1/x^2 grows even faster, making 2(1/x2) 2^{(1/x^2)} grow much larger than 5(1/x) 5^{(1/x)} .

This results in an indeterminate form: ∞∞ \frac{\infty}{\infty} .

Step 3: Apply logarithms


To simplify the expression, we take the natural logarithm. Define:

L=lim⁡x→05(1/x)2(1/x2)
L = \lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}}


Taking the natural logarithm of both sides:

ln⁡L=lim⁡x→0(ln⁡5(1/x)−ln⁡2(1/x2))
\ln L = \lim_{x \to 0} \left( \ln 5^{(1/x)} - \ln 2^{(1/x^2)} \right)


Simplifying further:

ln⁡L=lim⁡x→0(ln⁡5x−ln⁡2x2)
\ln L = \lim_{x \to 0} \left( \frac{\ln 5}{x} - \frac{\ln 2}{x^2} \right)




This limit involves terms that grow at different rates as x→0 x \to 0 , making it useful to find a common denominator to better analyze their behavior.

Step 4: Get a common denominator


The least common denominator between x x and x2 x^2 is x2 x^2 . Rewrite each term with this common denominator:

ln⁡5x=ln⁡5⋅xx2,ln⁡2x2=ln⁡2x2.
\frac{\ln 5}{x} = \frac{\ln 5 \cdot x}{x^2}, \quad \frac{\ln 2}{x^2} = \frac{\ln 2}{x^2}.


Now, combine the terms:

ln⁡5⋅x−ln⁡2x2.
\frac{\ln 5 \cdot x - \ln 2}{x^2}.

Step 5: Evaluate the limit


As x→0 x \to 0 , the term ln⁡5⋅x \ln 5 \cdot x vanishes, leaving:

lim⁡x→0(ln⁡5⋅x−ln⁡2x2)=lim⁡x→0−ln⁡2x2.
\lim_{x \to 0} \left( \frac{\ln 5 \cdot x - \ln 2}{x^2} \right) = \lim_{x \to 0} \frac{-\ln 2}{x^2}.


Since the numerator is a negative constant and the denominator grows large, the limit is:

lim⁡x→0−ln⁡2x2=−∞.
\lim_{x \to 0} \frac{-\ln 2}{x^2} = -\infty.


Step 6: Evaluate the original limit


Since:

ln⁡L→−∞  ⟹  L→0
\ln L \to -\infty \implies L \to 0


Step 5: Conclusion


The limit is:

lim⁡x→05(1/x)2(1/x2)=0
\boxed{\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}} = 0}


Prof's perspective


The professor likely asked this question to assess your understanding of indeterminate forms and your ability to apply logarithmic transformations effectively. This problem reinforces the idea that in calculus, growth rates of different functions (like exponential terms) are crucial for evaluating limits.
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lim⁡x→05(1/x)2(1/x2)=0
\boxed{\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}} = 0}