2E: Review Exercises

The following problems are collected from past exams.

Exercise $\PageIndex{1}$: Limits

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Find

1. $\displaystyle \lim\limits_{x \rightarrow -2} x^3+6x^2-16$

    

2. $\displaystyle \lim\limits_{x \rightarrow 0} \pi^2$

    

3. $\displaystyle \lim\limits_{x \rightarrow -\infty} \frac{\sqrt{3x^2+4x-1}}{3-x}$

    

4. $\displaystyle \lim\limits_{x \rightarrow 4} \frac{x^2+9}{x^2-1}$

    

5. $\displaystyle \lim\limits_{x \rightarrow 4} \frac{x^2-16}{x^2+x-20}$

    

6. $\displaystyle \lim\limits_{x \rightarrow 0} \frac{x^2+2x}{x-2x^2}$

    

7. $\displaystyle \lim\limits_{x \rightarrow 1} \frac{1-x^2}{x^2+5x-6}$

    

8. $\displaystyle \lim\limits_{x \rightarrow 3} \frac{x^3-27}{x-3}$

    

9. $\displaystyle \lim\limits_{x \rightarrow 4^-} \frac{x-4}{|x-4|}$

    

10. $\displaystyle \lim\limits_{x \rightarrow \infty} \frac{\sqrt{x^2-4}}{2x}$

     

11. $\displaystyle \lim\limits_{x \to 5}\displaystyle \frac{\sqrt{x-5}}{x}$

     

Answer

$$0, \pi^2, \sqrt{3}, 5/3, 8/9, 2, -2/7, 27, -1, 1/2, 0$$

Solution:

1. $\displaystyle \lim\limits_{x \rightarrow -2} x^3+6x^2-16$

Input x = -2, tthen,

$\displaystyle \lim\limits_{x \rightarrow -2} x^3+6x^2-16$= $(-2)^3+6(2)^2-16$= $-8+24-16$= 0

2. $\displaystyle \lim\limits_{x \rightarrow 0} \pi^2$

$\pi^2$ is just a number!! Therefore:

= $\pi^2$

3. $\displaystyle \lim\limits_{x \rightarrow -\infty} \frac{\sqrt{3x^2+4x-1}}{3-x}$

Because x is approaching $-\infty$, consider only the highest degree polynomials of the numerator and denominator:

= $\displaystyle \lim\limits_{x \rightarrow -\infty} \frac{\sqrt{3x^2}}{x}$

Divide the numerator and denominator by $\sqrt{x^2}$:

=$\displaystyle \lim\limits_{x \rightarrow -\infty} \frac{\frac{\sqrt{3x^2}}{\sqrt{x^2}}}{\frac{x}{x}}$

Simplify and solve:

=$\displaystyle \lim\limits_{x \rightarrow -\infty} \frac{\sqrt{3}}{1}$

= $\sqrt{3}$

4. $\displaystyle \lim\limits_{x \rightarrow 4} \frac{x^2+9}{x^2-1}$

Because the only domain restrictions for this function are x=1 and x=-1, you may input x=4 and solve:

= $\displaystyle \frac{(4)^2+9}{(4)^2-1}$

= $\frac{25}{15}$

= $\frac{5}{3}$

5. $\displaystyle \lim\limits_{x \rightarrow 4} \frac{x^2-16}{x^2+x-20}$

Due to the fact that one of the domain restrictions for this function is x=4, both the numerator and denominator must be factored in order to answer:

= $\displaystyle \lim\limits_{x \rightarrow 4} \frac{(x-4)(x+4)}{(x-4)(x+5)}$

Now the (x-4) from the numerator and the (x-4) from the denominator maybe be divided to equal 1:

= $\displaystyle \lim\limits_{x \rightarrow 4} \frac{x+4}{x+5}$

Now, because the limit is no longer a domain restriction, 4 may be input for x:

= $\displaystyle \frac{4+4}{4+5}$

= $\frac{8}{9}$

6. $\displaystyle \lim\limits_{x \rightarrow 0} \frac{x^2+2x}{x-2x^2}$

Due to the fact that one of the domain restrictions for this function is x=0, both the numerator and denominator must be factored in order to answer:

= $\displaystyle \lim\limits_{x \rightarrow 0} \frac{x(x+2)}{x(1-2x)}$

Now the x from the numerator and the x from the denominator maybe be divided to equal 1:

= $\displaystyle \lim\limits_{x \rightarrow 0} \frac{x+2}{1-2x}$

Now, because the limit is no longer a domain restriction, 0 may be input for x:

= $\displaystyle \frac{(0)+2}{1-2(0)}$

= $\frac{2}{1} = 2$

7. $\displaystyle \lim\limits_{x \rightarrow 1} \frac{1-x^2}{x^2+5x-6}$

Due to the fact that one of the domain restrictions for this function is x=1, both the numerator and denominator must be factored in order to answer:

= $\displaystyle \lim\limits_{x \rightarrow 1} \frac{(1-x)(1+x)}{(x+6)(x-1)}$

Looking at this, it doesn't seem as though any of the terms divide to equal 1, however you can make one of the terms in the numerator the same by factoring out a -1:

= $\displaystyle \lim\limits_{x \rightarrow 1} \frac{-(-1+x)(1+x)}{(x+6)(x-1)}$

= $\displaystyle \lim\limits_{x \rightarrow 1} \frac{-(x-1)(1+x)}{(x+6)(x-1)}$

Now the (x-1) from the numerator and the (x-1) from the denominator maybe be divided to equal 1:

= $\displaystyle \lim\limits_{x \rightarrow 1} \frac{-(1+x)}{(x+6)}$

Now, because the limit is no longer a domain restriction, 1 may be input for x:

= $\displaystyle \frac{-(1+(1))}{((1)+6)}$

= $\frac{-2}{7}$

8. $\displaystyle \lim\limits_{x \rightarrow 3} \frac{x^3-27}{x-3}$

Due to the fact that one of the domain restrictions for this function is x=3, the numerator must be factored in order to answer:

= $\displaystyle \lim\limits_{x \rightarrow 3} \frac{(x-3)(x^2+3x+3^2}{x-3}$

Now the (x-3) from the numerator and the (x-3) from the denominator maybe be divided to equal 1:

= $\displaystyle \lim\limits_{x \rightarrow 3} x^2+3x+3^2$

Now, because the limit is no longer a domain restriction, 3 may be input for x:

= $\displaystyle (3)^2+3(3)+3^2$

= $\displaystyle 9+9+9 = 27$

9. $\displaystyle \lim\limits_{x \rightarrow 4^-} \frac{x-4}{|x-4|}$

Due to absolute value properties $|x-4| = x$ if $x \geq 0$ or $-x$ if $x < 0$.

Because $x-4 < 0$ when x approaches $4^-$ that means that $|x-4| = -x+4$:

= $\displaystyle \lim\limits_{x \rightarrow 4^-} \frac{x-4}{-x+4}$

To make the numerator and denominator look similar, a -1 may be factored out of the numerator:

= $\displaystyle \lim\limits_{x \rightarrow 4^-} \frac{-(-x+4)}{-x+4}$

Now the (x-3) from the numerator and the (x-3) from the denominator maybe be divided to equal 1:

= $\displaystyle \lim\limits_{x \rightarrow 4^-} -1$

Because there are no variables left, the limit is simply equal to -1

10. $\displaystyle \lim\limits_{x \rightarrow \infty} \frac{\sqrt{x^2-4}}{2x}$

Because the limit is approaching $\infty$, only the highest degree polynomials from the numerator and the denominator need to be considered:

= $\displaystyle \lim\limits_{x \rightarrow \infty} \frac{\sqrt{x^2}}{2x}$

This can now be simplified:

= $\displaystyle \lim\limits_{x \rightarrow \infty} \frac{x}{2x}$

= $\displaystyle \lim\limits_{x \rightarrow \infty} \frac{1}{2}$

Because there are no variables left, the limit is simply equal to 1/2

11. $\displaystyle \lim\limits_{x \to 5}\displaystyle \frac{\sqrt{x-5}}{x}$

Because the only domain restriction is when x=0, 5 may simply be inputted for x:

= $\displaystyle \frac{\sqrt{(5)-5}}{5}$

This may now be simplified:

= $\displaystyle \frac{\sqrt{0}}{5}$

= $0$

Exercise $\PageIndex{2}$: Trigonometry Limits

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1. $\lim_{t \to 0} \displaystyle \frac{\sin (t^2) +t^2}{2t \sin (t)}$
2. $\displaystyle{\lim_{t \rightarrow 0} \frac{\sin(2t)}{t\cos(t)}}$
3. $\lim_{x \rightarrow 0^+}x^2 \csc^2x$
4. $\lim_{t \to 0} \displaystyle \frac{\sqrt{t} \sin (\sqrt{t}) }{-4t + \sin (t)}$
5. $\lim_{x \to 0} \displaystyle \frac{\tan^2(\frac{x}{3})}{4x^2}$
6. $\lim_{x \to \pi} \displaystyle \frac{\sin(\pi -x)}{\pi-x}$

Answer

$$1,2,1,\frac{-1}{3},\frac{1}{36},1$$

Solutions:

Add texts here. Do not delete this text first.

1. $\lim_{t \to 0} \displaystyle \frac{\sin (t^2) +t^2}{2t \sin (t)}$

Factor out $t^2$ from the numerator and multiply the denominator by $\frac{t}{t}$:

= $\lim_{t \to 0} \displaystyle \frac{t^2(\frac{\sin (t^2)}{t^2} + 1)}{2t^2 \frac{\sin (t)}{t}}$

Use the trig limit identity $\lim_{t \to 0} \displaystyle \frac{sin(x)}{x} = 1$:

= $\lim_{t \to 0} \displaystyle \frac{t^2(1 + 1)}{2t^2}$

= $\lim_{t \to 0} \displaystyle \frac{2t^2}{2t^2}$

= $1$

2. $\displaystyle{\lim_{t \rightarrow 0} \frac{\sin(2t)}{t\cos(t)}}$

Multiply the limit by $\frac{2}{2}$:

= $\displaystyle{\lim_{t \rightarrow 0} \frac{\sin(2t)}{t\cos(t)} \frac{2}{2}}$

= $\displaystyle{2\lim_{t \rightarrow 0} \frac{\sin(2t)}{2t\cos(t)}}$

Use the trig limit identity $\lim_{t \to 0} \displaystyle \frac{sin(x)}{x} = 1$:

= $\displaystyle{2\lim_{t \rightarrow 0} \frac{1}{\cos(t)}}$

Input 0 for t:

= $\displaystyle{2\frac{1}{\cos(0)}}$

= $\displaystyle{2\frac{1}{1}}$

= $2$

3. $\lim_{x \rightarrow 0^+}x^2 \csc^2x$

Use trig property $\csc(x) = \frac{1}{\sin^2(x)}$:

= $\lim_{x \rightarrow 0^+}x^2 \frac{1}{\sin^2(x)}$

Use the trig limit identity $\lim_{t \to 0} \displaystyle \frac{sin(x)}{x} = 1$ (Notice how the inverse is also equal to 1):

= $\lim_{x \rightarrow 0^+}x \frac{1}{\sin(x)}$

Use the trig limit identity $\lim_{t \to 0} \displaystyle \frac{sin(x)}{x} = 1$ again:

= $\lim_{x \rightarrow 0^+}1$

=1

4. $\lim_{t \to 0} \displaystyle \frac{\sqrt{t} \sin (\sqrt{t}) }{-4t + \sin (t)}$

Multiply the numerator by $\frac{\sqrt{t}}{\sqrt{t}}$ and the factor the denominator by $t$:

=$\lim_{t \to 0} \displaystyle \frac{\frac{\sqrt{t}}{\sqrt{t}}(\sqrt{t} \sin (\sqrt{t})) }{t(-4 + \frac{\sin (t)}{t})}$

Use the trig limit identity $\lim_{t \to 0} \displaystyle \frac{sin(x)}{x} = 1$:

=$\lim_{t \to 0} \displaystyle \frac{1t}{t(-4 + 1)}$

Simplify:

=$\lim_{t \to 0} \displaystyle \frac{1}{-3}$

Therefore:

= $\frac{-1}{3}$

5. $\lim_{x \to 0} \displaystyle \frac{\tan^2(\frac{x}{3})}{4x^2}$

Use trig identity $\tan(x)=\frac{\sin(x)}{\cos(x)}$:

=$\frac{1}{4}\lim_{x \to 0} \displaystyle \frac{\frac{\sin^2\frac{x}{3}}{\cos^2\frac{x}{3}}}{x^2}$

Multiply the denominator by $\frac{3}{3}$ and simplify the fraction:

=$\frac{1}{4}\lim_{x \to 0} \displaystyle \frac{\sin^2\frac{x}{3}}{\frac{3x(x)}{3}\cos^2\frac{x}{3}}$

Use the trig limit identity $\lim_{t \to 0} \displaystyle \frac{sin(x)}{x} = 1$:

=$\frac{1}{12}\lim_{x \to 0} \displaystyle \frac{\sin \frac{x}{3}}{(x) \cos^2\frac{x}{3}}$

Multiply the denominator by $\frac{3}{3}$:

=$\frac{1}{12}\lim_{x \to 0} \displaystyle \frac{\sin \frac{x}{3}}{\frac{3x}{3} \cos^2\frac{x}{3}}$

Use the trig limit identity $\lim_{t \to 0} \displaystyle \frac{sin(x)}{x} = 1$:

=$\frac{1}{36}\lim_{x \to 0} \displaystyle \frac{1}{\cos^2\frac{x}{3}}$

Input 0 for x and solve:

=$\frac{1}{36}\displaystyle \frac{1}{\cos^2(0)}$

=$\frac{1}{36}\displaystyle \frac{1}{1}$

= $\frac{1}{36}$

6. $\lim_{x \to \pi} \displaystyle \frac{\sin(\pi -x)}{\pi-x}$

Let $t = \pi - x$ and find $\lim_{t \to 0}$:

= $\lim_{t \to 0} \displaystyle \frac{\sin(t))}{t}$

Use the trig limit identity $\lim_{t \to 0} \displaystyle \frac{sin(x)}{x} = 1$:

= $\lim_{t \to 0} \displaystyle 1$

Change parameters back to x:

= $\lim_{x \to \pi} \displaystyle 1$

= $1$

Exercise $\PageIndex{3}$: Limits

Find the following limits:

1. $$\lim_{x \to -\infty} \displaystyle \frac{3x+4}{\sqrt{x^2-x-1}}$$
2. $$\lim_{x \to -2^{-}} \displaystyle \frac{|x+2|}{x+2}$$
3. $$\lim_{x \to -2} \displaystyle \frac{3-x}{x+2}$$
4. $$\lim_{x \to 2} \sqrt{2-x}-x$$
5. $$\lim_{x \rightarrow -\infty}\frac{2x-1}{\sqrt{3x^2+x+6}}$$
6. $$\lim_{x \rightarrow 1^+}{\sqrt{-x^2+x}}$$
7. $$\lim_{h \rightarrow 0}\frac{(3+h)^{-1}-3^{-1}}{h}$$
8. $$\lim_{x \rightarrow 0}\frac{|3x-1|-|3x+1|}{x}$$
9. $$\lim_{t\rightarrow 0}\left(\frac{1}{t \sqrt{t+1}}-\frac{1}{t}\right)$$
10. $$\displaystyle \lim_{x \rightarrow 0}\frac{\sin x}{\tan x}$$

Answer

$$-3, 1, DNE, -1, \frac{-2}{\sqrt{3}}, 0, \frac{-1}{9}, -6, \frac{-1}{2}, 1$$

Exercise $\PageIndex{4}$: Asymptotes

Find all vertical asymptoe(s) (if any) and all horizontal asymptote(s) (if any) for the function
$$\displaystyle{f(x)=\frac{x^2-5x+6}{x^2-4x+3}}$$

Justify your work using limits.

Answer

The vertical asymptote is $x=1$ and the horizontal asymptote is $y=1$.

Exercise $\PageIndex{5}$

1. $\lim\limits_{x \to 0}\frac{\ln (1+x)}{x}$
2. $\lim\limits_{x\to0}\frac{\sin x}{x}$, where x is measured in degrees not radians.

Answer

Coming soon.

Contributors and Attributions

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/

Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)