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<x<4 \\ x^2 +2 & \mbox{if }x \geq 4. \end{cases}
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.


Last modified on 22 June 2013, at 09:17