# Calculus/Limits/Solutions

## Basic Limit ExercisesEdit

1. $\lim_{x\to 2} (4x^2 - 3x+1)$

Since this is a polynomial, two can simply be plugged in. This results in $4(4)-2(3)+1=16-6+1=\mathbf{11}$

2. $\lim_{x\to 5} (x^2)$

$5^2=\mathbf{25}$

## One-Sided LimitsEdit

Evaluate the following limits or state that the limit does not exist.

3. $\lim_{x\to 0^-} \frac{x^3+x^2}{x^3+2x^2}$

Factor as $\frac{x^2}{x^2}\frac{x+1}{x+2}$. In this form we can see that there is a removable discontinuity at x=0 and that the limit is $\mathbf{\frac{1}{2}}$

4. $\lim_{x\to 7^-} |x^2+x| -x$

$|7^2+7|-7 = \mathbf{49}$

5. $\lim_{x\to -1^+} \sqrt{1-x^2}$

$\sqrt{1-x^2}$ is defined if $x^2<1$, so the limit is $\sqrt{1-1^2}=\mathbf{0}$

6. $\lim_{x\to -1^-} \sqrt{1-x^2}$

$\sqrt{1-x^2}$ is not defined if $x^2>1$, so the limit does not exist.

## Two-Sided LimitsEdit

Evaluate the following limits or state that the limit does not exist.

7. $\lim_{x \to -1} \frac{1}{x-1}$

$\mathbf{-\frac{1}{2}}$

8. $\lim_{x\to 4} \frac{1}{x-4}$

$\lim_{x\to 4^-} \frac{1}{x-4}=-\infty$
$\lim_{x\to 4^+} \frac{1}{x-4}=+\infty$
The limit does not exist.

9. $\lim_{x\to 2} \frac{1}{x-2}$

$\lim_{x\to 2^-} \frac{1}{x-2}=-\infty$
$\lim_{x\to 2^+} \frac{1}{x-2}=+\infty$
The limit does not exist.

10. $\lim_{x\to -3} \frac{x^2 - 9}{x+3}$

$\lim_{x\to -3} \frac{(x+3)(x-3)}{x+3} = \lim_{x\to -3} x-3 = -3-3=\mathbf{-6}$

11. $\lim_{x\to 3} \frac{x^2 - 9}{x-3}$

$\lim_{x\to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3} x+3 = 3+3 = \mathbf{6}$

12. $\lim_{x\to -1} \frac{x^2+2x+1}{x+1}$

$\lim_{x\to -1} \frac{(x+1)(x+1)}{x+1} = \lim_{x\to -1} x+1 = -1+1 = \mathbf{0}$

13. $\lim_{x\to -1} \frac{x^3+1}{x+1}$

$\lim_{x\to -1} \frac{(x^2-x+1)(x+1)}{x+1} = \lim_{x\to -1} x^2-x+1 = (-1)^2-(-1)+1 = 1+1+1 = \mathbf{3}$

14. $\lim_{x\to 4} \frac{x^2 + 5x-36}{x^2 - 16}$

$\lim_{x\to 4} \frac{(x-4)(x+9)}{(x-4)(x+4)} = \lim_{x\to 4} \frac{x+9}{x+4} = \frac{4+9}{4+4} = \mathbf{\frac{13}{8}}$

15. $\lim_{x\to 25} \frac{x-25}{\sqrt{x}-5}$

$\lim_{x\to 25} \frac{(\sqrt{x}-5)(\sqrt{x}+5)}{\sqrt{x}-5} = \lim_{x\to 25} \sqrt{x}+5) = \sqrt{25}+5) = 5+5 = \mathbf{10}$

16. $\lim_{x\to 0} \frac{\left|x\right|}{x}$

$\lim_{x\to 0^-} \frac{\left|x\right|}{x} = \lim_{x\to 0^-} \frac{-x}{x} = \lim_{x\to 0^-} -1 = -1$
$\lim_{x\to 0^+} \frac{\left|x\right|}{x} = \lim_{x\to 0^+} \frac{x}{x} = \lim_{x\to 0^+} 1 = 1$
The limit does not exist.

17. $\lim_{x\to 2} \frac{1}{(x-2)^2}$

As $x$ approaches $2$, the denominator will be a very small positive number, so the whole fraction will be a very large positive number. Thus, the limit is $\mathbf{\infty}$.

18. $\lim_{x\to 3} \frac{\sqrt{x^2+16}}{x-3}$

As $x$ approaches $3$, the numerator goes to 5 and the denominator goes to 0. Depending on whether you approach $3$ from the left or the right, the denominator will be either a very small negative number, or a very small positive number. So the limit from the left is $-\infty$ and the limit from the right is $+\infty$. Thus, the limit does not exist.

19. $\lim_{x\to -2} \frac{3x^2-8x -3}{2x^2-18}$

$\frac{3(-2)^2-8(-2) -3}{2(-2)^2-18} = \frac{3(4)+16-3}{2(4)-18} = \frac{12+16-3}{8-18} = \frac{25}{-10} = \mathbf{-\frac{5}{2}}$

20. $\lim_{x\to 2} \frac{x^2 + 2x + 1}{x^2-2x+1}$

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

21. $\lim_{x\to 3} \frac{x+3}{x^2-9}$

$\lim_{x\to 3} \frac{x+3}{(x+3)(x-3)} = \lim_{x\to 3} \frac{1}{x-3}$
$\lim_{x\to 3^{-}} \frac{1}{x-3} = -\infty$
$\lim_{x\to 3^{+}} \frac{1}{x-3} = +\infty$
The limit does not exist.

22. $\lim_{x\to -1} \frac{x+1}{x^2+x}$

$\lim_{x\to -1} \frac{x+1}{x(x+1)} = \lim_{x\to -1} \frac{1}{x} = \frac{1}{-1} = \mathbf{-1}$

23. $\lim_{x\to 1} \frac{1}{x^2+1}$

$\frac{1}{1^2+1} = \frac{1}{1+1} = \mathbf{\frac{1}{2}}$

24. $\lim_{x\to 1} x^3 + 5x - \frac{1}{2-x}$

$1^3 + 5(1) - \frac{1}{2-1} = 1 + 5 - \frac{1}{1} = 6 - 1 = \mathbf{5}$

25. $\lim_{x\to 1} \frac{x^2-1}{x^2+2x-3}$

$\lim_{x\to 1} \frac{(x-1)(x+1)}{(x-1)(x+3)} = \lim_{x\to 1} \frac{x+1}{x+3} = \frac{1+1}{1+3} = \frac{2}{4} = \mathbf{\frac{1}{2}}$

26. $\lim_{x\to 1} \frac{5x}{x^2+2x-3}$

Notice that as $x$ approaches $1$, the numerator approaches $5$ while the denominator approaches $0$. However, if you approach from below, the denominator is negative, and if you approach from above, the denominator is positive. So the limits from the left and right will be $-\infty$ and $+\infty$ respectively. Thus, the limit does not exist.

## Limits to InfinityEdit

Evaluate the following limits or state that the limit does not exist.

27. $\lim_{x\to \infty} \frac{-x + \pi}{x^2 + 3x + 2}$

This rational function is bottom-heavy, so the limit is $\mathbf{0}$.

28. $\lim_{x\to -\infty} \frac{x^2+2x+1}{3x^2+1}$

This rational function has evenly matched powers of $x$ in the numerator and denominator, so the limit will be the ratio of the coefficients, i.e. $\mathbf{\frac{1}{3}}$.

29. $\lim_{x\to -\infty} \frac{3x^2 + x}{2x^2 - 15}$

Balanced powers in the numerator and denominator, so the limit is the ratio of the coefficients, i.e. $\mathbf{\frac{3}{2}}$.

30. $\lim_{x\to -\infty} 3x^2-2x+1$

This is a top-heavy rational function, where the exponent of the ratio of the leading terms is $2$. Since it is even, the limit will be $\mathbf{\infty}$.

31. $\lim_{x\to \infty} \frac{2x^2-32}{x^3-64}$

Bottom-heavy rational function, so the limit is $\mathbf{0}$.

32. $\lim_{x\to \infty} 6$

This is a rational function, as can be seen by writing it in the form $\frac{6x^0}{1x^0}$. Since the powers of $x$ in the numerator and denominator are evenly matched, the limit will be the ratio of the coefficients, i.e. $\mathbf{6}$.

33. $\lim_{x\to \infty} \frac{3x^2 +4x}{x^4+2}$

Bottom-heavy, so the limit is $\mathbf{0}$.

34. $\lim_{x\to -\infty} \frac{2x+3x^2+1}{2x^2+3}$

Evenly matched highest powers of $x$ in the numerator and denominator, so the limit will be the ratio of the corresponding coefficients, i.e. $\mathbf{\frac{3}{2}}$.

35. $\lim_{x\to -\infty} \frac{x^3-3x^2+1}{3x^2+x+5}$

Top-heavy rational function, where the exponent of the ratio of the leading terms is $1$, so the limit is $\mathbf{-\infty}$.

36. $\lim_{x\to \infty} \frac{x^2+2}{x^3-2}$

Bottom-heavy, so the limit is $\mathbf{0}$.

## Limits of Piecewise FunctionsEdit

Evaluate the following limits or state that the limit does not exist.

37. Consider the function

$f(x) = \begin{cases} (x-2)^2 & \mbox{if }x<2 \\ x-3 & \mbox{if }x\geq 2. \end{cases}$
a. $\lim_{x\to 2^-}f(x)$

$(2-2)^2 = \mathbf{0}$

b. $\lim_{x\to 2^+}f(x)$

$2-3 = \mathbf{-1}$

c. $\lim_{x\to 2}f(x)$

Since the limits from the left and right don't match, the limit does not exist.

38. Consider the function

$g(x) = \begin{cases} -2x+1 & \mbox{if }x\leq 0 \\ x+1 & \mbox{if }0
a. $\lim_{x\to 4^+} g(x)$

$4^2+2 = 16+2 = \mathbf{18}$

b. $\lim_{x\to 4^-} g(x)$

$4+1 = \mathbf{5}$

c. $\lim_{x\to 0^+} g(x)$

$0+1 = \mathbf{1}$

d. $\lim_{x\to 0^-} g(x)$

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

e. $\lim_{x\to 0} g(x)$

Since the left and right limits match, the overall limit is also $\mathbf{1}$.

f. $\lim_{x\to 1} g(x)$

$1+1 = \mathbf{2}$

39. Consider the function

$h(x) = \begin{cases} 2x-3 & \mbox{if }x<2 \\ 8 & \mbox{if }x=2 \\ -x+3 & \mbox{if } x>2. \end{cases}$
a. $\lim_{x\to 0} h(x)$

$2(0)-3 = \mathbf{-3}$

b. $\lim_{x\to 2^-} h(x)$

$2(2)-3 = 4-3 = \mathbf{1}$

c. $\lim_{x\to 2^+} h(x)$

$-(2)+3 = \mathbf{1}$

d. $\lim_{x\to 2} h(x)$

Since the limits from the right and left match, the overall limit is $\mathbf{1}$. Note that in this case, the limit at 2 does not match the function value at 2, so the function is discontinuous at this point, hence the function is nondifferentiable at this point as well.