1.4 Computing Limits

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Compute limits using algebraic techniques.

Computing Limits

We have seen examples of limit problems where plugging in the terminal $x$ value both led to reasonable answers and meaningless forms. In this section we will work with examples in which plugging in the terminal $x$ value initially yields a meaningless expression, but after performing algebraic manipulations on the function in the problem, plugging in the terminal value yields a solution to the problem.

Factor and Cancel

Compute the limit: $\lim _{x \to 4} \frac {x^2 - 16}{x^2 - 4x}.$

Plugging in the terminal value, $x=4$ , yields the indeterminate form $0/0$ . To find this limit analytically, we will factor the numerator and denominator and simplify the fraction. In the numerator we have a difference of squares and such a form always factors in the following way: $a^2 - b^2 = (a+b)(a-b).$ Applying this general formula to our example, we get $x^2 - 16 = (x+4)(x-4).$ Next, in the denominator, we can factor out a common factor of $x$ from each of the terms: $x^2 - 4x = x(x-4).$ With these factorizations we can simplify the fraction and find the limit:

$\begin{align*} \lim_{x \to 4} \frac{x^2 - 16}{x^2 - 4x} &= \lim_{x \to 4} \frac{(x+4)(x-4)}{x(x-4)} \enspace & \text{(factoring)} \\ &= \lim_{x \to 4} \frac{x+4}{x} & \text{(canceling)} \\ &= \frac84 & \\ &= 2. & \end{align*}$

Hence, $\lim _{x \to 4} \frac {x^2 - 16}{x^2 - 4x} = 2.$ This example used the factor and cancel method.

Compute the limit: $\lim _{x \to 3} \frac {x^2 - 9}{x^2 - 3x} = \answer {2}.$

When you plug in $x = 3$ , you get $\frac 00$

Factor the numerator and the denominator

Difference of squares: $a^2 - b^2 = (a+b)(a-b)$

Common factor: $ab \pm ac = a(b \pm c)$

Compute the limit: $\lim _{x \to 2} \frac {x^2 + 3x - 10}{x^3 - 8}.$
Plugging in the terminal value $x = 2$ yields the indeterminate form $\frac 00$ . The numerator factors by finding two numbers which multiply to give $-10$ and add to give $+3$ . The two numbers are $-2$ and $+5$ , hence the factorization is $x^2 + 3x - 10 = (x-2)(x+5).$ In the denominator, we have a difference of cubes and we use the formula: $a^3 - b^3 = (a-b)(a^2+ab +b^2).$ Applying this to our example gives $x^3 - 8 = (x-2)(x^2 + 2x + 4).$ With these factorizations, we can simplify the fraction and find the limit: $\begin{align*} \lim_{x \to 2} \frac{x^2 + 3x - 10}{x^3 - 8} &= \lim_{x \to 2}\frac{(x-2)(x+5)}{(x-2)(x^2 + 2x + 4)} \enspace & \text{(factoring)} \\ &= \lim_{x \to 2} \frac{x+5}{x^2 + 2x + 4} & \text{(canceling)} \\ &= \frac{2+5}{2^2 + (2\cdot 2) + 4} & \text{(plugging in)} \\ &= \frac{7}{12}. \end{align*}$

Hence, $\lim _{x \to 2} \frac {x^2 + 3x - 10}{x^3 - 8} = \frac {7}{12}.$

Compute the following limit which we investigated numerically in the previous section: $\lim _{x \to 1} \frac {x^3 - 1}{x^2 -1}.$

When you plug in $x = 1$ , you get $\frac 00$

Factor the numerator and the denominator

Difference of cubes: $a^3 - b^3 = (a-b)(a^2 + ab +b^2)$

Difference of squares: $a^2 - b^2 = (a-b)(a+b)$

The value of the limit is $\answer [given]{\frac 32}$

Compute the limit: $\lim _{x \to 3} \frac {x^3 - 27}{x^2 -2x - 3}.$

When you plug in $x = 3$ , you get $\frac 00$

Factor the numerator and the denominator

Difference of cubes: $a^3 - b^3 = (a-b)(a^2 + ab +b^2)$

The value of the limit is $\answer [given]{\frac {27}{4}}$

Compute the limit: $\lim _{x \to 4} \frac {4x^2 - 19x + 12}{6x^2 -19x -20}.$
Plugging in the terminal value $x = 4$ yields the indeterminate form $\frac 00$ . To factor the numerator and denominator in this case, we will use the important fact that for a polynomial $p(x)$ and a number $a$ , $\text {If} \quad p(a) = 0 \quad \text {then} \quad x-a \quad \text {is a factor of} \; p(x).$ And once we know one factor of a polynomial, it is usually much easier to find the other. Since plugging in $x=4$ gave $0$ in both the numerator and the denominator, both polynomials have $x-4$ as a factor. In the numerator, the factorization is $(x-4)(4x-3)$ and in the denominator, the factorization is $(x-4)(6x+5)$ .

With these factorizations, we can simplify the fraction and find the limit:

$\begin{align*} \lim_{x \to 4} \frac{4x^2 - 19x + 12}{6x^2 -19x -20} &= \lim_{x \to 4}\frac{(x-4)(4x-3)}{(x-4)(6x + 5)} \enspace & \text{(factoring)} \\ &= \lim_{x \to 4} \frac{4x-3}{6x + 5} & \text{(canceling)} \\ &= \frac{16-3}{24+5} & \text{(plugging in)}\\ &= \frac{13}{29}. \end{align*}$

Hence, $\lim _{x \to 4} \frac {4x^2 -19x +12}{6x^2 -19x -20} = \frac {13}{29}.$

Compute the limit: $\lim _{x \to 2} \frac {3x^2 -8x + 4}{2x^2 + x - 10}.$

When you plug in $x = 2$ , you get $\frac 00$

Factor the numerator and the denominator

If $p(a) = 0$ then $(x-a)$ is a factor

$x-2$ is a factor of both the numerator and the denominator

The value of the limit is $\answer [given]{4/9}$

Compute the limit: $\lim _{x \to -2} \frac {2x^2 + x - 6}{x^2 + 5x + 6}.$

When you plug in $x = -2$ , you get $\frac 00$

Factor the numerator and the denominator

If $p(a) = 0$ then $(x-a)$ is a factor

$x+2$ is a factor of both the numerator and the denominator

The value of the limit is $\answer [given]{-7}$

Conjugate Radicals

The expressions $\sqrt u + \sqrt v \quad \text {and} \quad \sqrt u - \sqrt v$ are called conjugate radicals. When we multiply conjugate radicals using the difference of squares formula, $(a+b)(a-b) = a^2 - b^2$ , we get an expression that is free of radicals:

$(\sqrt u + \sqrt v)(\sqrt u - \sqrt v) = (\sqrt u)^2 - (\sqrt v)^2 = u - v.$

We will now take advantage of this technique to find limits.

Compute the limit: $\lim _{x \to 4} \frac {\sqrt {x} - 2}{x-4}.$

To solve this limit problem, we will use the conjugate radical of the numerator, which is $\sqrt {x} + 2$ . In order to maintain the value of the expression given in the problem, we multiply by the conjugate radical over itself (which is equal to one):

$\begin{align*} \lim_{x \to 4} \frac{\sqrt{x}- 2}{x-4} &= \lim_{x \to 4} \frac{\sqrt{x} -2}{x-4}\cdot \frac{\sqrt{x} +2}{\sqrt{x}+2} \\ &= \lim_{x \to 4} \frac{x-4}{(x-4)(\sqrt{x}+2)} \\ &= \lim_{x \to 4}\frac{1}{\sqrt{x}+2} \\ &= \frac 14. \end{align*}$

In the last step, we plugged in the terminal value $x=4$ to get the final answer $\frac 14$ .

Compute the limit: $\lim _{x \to 9} \frac {\sqrt {x}-3}{x-9}$

When you plug in $x = 9$ , you get $\frac 00$

Multiply by the conjugate radical

$\sqrt a + \sqrt b$ and $\sqrt a - \sqrt b$ are conjugates

Use the difference of squares formula in the numerator: $(a-b)(a+b) = a^2 - b^2$

The value of the limit is $\answer [given]{1/6}$

Compute the limit: $\lim _{x \to 1} \frac {x- \sqrt {x}}{x-1}.$

To solve this limit problem, we will use the conjugate radical of the numerator, which is $x+ \sqrt {x}$ . In order to not change the value of the expression given in the problem, we multiply by the conjugate radical divided by itself (which is equal to one):

$\begin{align*} \lim_{x \to 1} \frac{x- \sqrt{x}}{x-1} &= \lim_{x \to 1} \frac{x- \sqrt{x}}{x-1}\cdot \frac{x+ \sqrt{x}}{x+ \sqrt{x}} \\ &= \lim_{x \to 1} \frac{x^2- x}{(x-1)(x+ \sqrt{x})} \\ &= \lim_{x \to 1} \frac{x(x-1)}{(x-1)(x+ \sqrt{x})} \\ &= \lim_{x \to 1}\frac{x}{x+ \sqrt{x}} \\ &= \frac 12. \end{align*}$

In the last step, we plugged in the terminal value $x=1$ to get the final answer $\frac 12$ .

Compute the limit: $\lim _{x \to 2} \frac {x- \sqrt {2x}}{x-2}$

When you plug in $x = 2$ , you get $\frac 00$

Multiply by the conjugate radical

$\sqrt a + \sqrt b$ and $\sqrt a - \sqrt b$ are conjugates

Use the difference of squares formula in the numerator: $(a-b)(a+b) = a^2 - b^2$

The value of the limit is $\answer [given]{\frac 12}$

Compute the limit: $\displaystyle {\lim _{x \to -3} \frac {x^2 - 9}{4 - \sqrt {13 -x}}}$ .

If we plug in $x = -3$ , the numerator and denominator are both 0. The conjugate of the radical expression in the denominator is $4 + \sqrt {13 -x}.$ To simplify our calculations a little bit, lets multiply the conjugates together separately: $(4 - \sqrt {13 -x})(4 + \sqrt {13 -x}) = 16 - (13-x) = 3+x = x+3.$ Now back to our problem: $\begin{align*} \lim_{x \to -3} \frac{x^2 - 9}{4 - \sqrt{13 -x}} &= \lim_{x \to -3} \left(\frac{x^2 - 9}{4 - \sqrt{13 -x}}\right) \cdot \left(\frac{4 + \sqrt{13 -x}}{4 + \sqrt{13 -x}}\right) \\ &=\lim_{x \to -3} \frac{(x^2 - 9)(4 + \sqrt{13 -x})}{x+3}\\ &=\lim_{x \to -3} \frac{(x+3)(x-3)(4 + \sqrt{13 -x})}{x+3} \\ &= \lim_{x \to -3} (x-3)(4 + \sqrt{13 -x})\\ &= (-3-3)(4 + \sqrt{13- (-3)}) \\ &= (-6)(4 + 4) \\ &= -48. \end{align*}$

Compute the limit: $\lim _{x \to -4} \frac {x^2 - 16}{3 - \sqrt {5 -x}}$

When you plug in $x = -4$ , you get $\frac 00$

Multiply by the conjugate radical

$\sqrt a + \sqrt b$ and $\sqrt a - \sqrt b$ are conjugates

Use the difference of squares formula in the denominator: $(a-b)(a+b) = a^2 - b^2$

$\sqrt a \cdot \sqrt a = a$

The value of the limit is $\answer [given]{-48}$

Compute the limit: $\lim _{x \to 0} \frac {4- \sqrt {x + 16}}{x}.$

When you plug in $x = 0$ , you get $\frac 00$

Multiply by the conjugate radical

$\sqrt a + \sqrt b$ and $\sqrt a - \sqrt b$ are conjugates

Use the difference of squares formula in the numerator: $(a-b)(a+b) = a^2 - b^2$

$\sqrt a \cdot \sqrt a = a$

The value of the limit is $\answer [given]{-\frac 18}$

Complex Fractions

We now consider examples involving fractions within fractions, called complex fractions.

Compute the limit: $\displaystyle {\lim _{x \to 4} \frac {\frac {1}{x} - \frac {1}{4}}{x-4}}.$

Plugging in $x=4$ yields the familiar $0/0$ indeterminate form. The algebra skills necessary to transform the function in the problem to one which allows plugging in $x=4$ involve subtraction and division of fractions. The rules are summarized as: $\frac {a}{b} - \frac {c}{d} = \frac {ad-bc}{bd} \quad \text {and} \quad \frac {\frac {a}{b}}{\frac {c}{d}} = \frac {ad}{bc}.$ Applying these to our problem, we get $\begin{align*} \lim_{x \to 4} \frac{\frac{1}{x} - \frac{1}{4}}{x-4} &= \lim_{x \to 4} \frac{\frac{4-x}{4x}}{x-4}\\ &=\lim_{x \to 4} \frac{4-x}{4x(x-4)}\\ &= \lim_{x \to 4} -\frac{1}{4x} \\ &= -\frac{1}{16}. \end{align*}$

It is important to note that $a-b$ and $b-a$ are opposites which cancel, leaving $-1$ : $\frac {a-b}{b-a} = -1.$ The the ratio of opposites is $-1$ .

Compute the limit: $\lim _{x \to 5} \frac {\frac {1}{x} - \frac {1}{5}}{x-5}.$

When you plug in $x = 5$ , you get $\frac 00$

Subtract the fractions in the numerator

$\frac {a}{b} \pm \frac {c}{d} = \frac {ad \pm bc}{bd}$ .

To divide, multiply by the reciprocal: $\frac {a}{b} \div \frac {c}{d} = \frac {a}{b} \cdot \frac {d}{c}$

Simplify the fraction by canceling

The value of the limit is $\answer [given]{-\frac {1}{25}}$

Compute the limit: $\lim _{x \to -2} \frac {\frac {2}{x-3} + \frac {x+4}{5}}{x^2 + 5x + 6}.$ Carefully plugging in $x=-2$ gives $\frac 00$ , so we begin simplifying the complex fraction. Since this function is bulky, let’s do the addition of fractions from the numerator separately: $\begin{align*} \frac{2}{x-3} + \frac{x+4}{5} &= \frac{10 + (x-3)(x+4)}{5(x-3)}\\ &= \frac{x^2 +x -2}{5(x-3)} \\ &= \frac{(x-1)(x+2)}{5(x-3)}. \end{align*}$

Now back to the original problem:

$\begin{align*} \lim_{x \to -2} \frac{\frac{2}{x-3} + \frac{x+4}{5}}{x^2 + 5x + 6} &= \lim_{x \to -2} \frac{\frac{(x-1)(x+2)}{5(x-3)}}{(x+2)(x+3)}\\ &= \lim_{x \to -2} \frac{(x-1)(x+2)}{5(x-3)(x+2)(x+3)} \\ &= \lim_{x \to -2} \frac{(x-1)}{5(x-3)(x+3)}\\ &= \frac{-3}{5(-5)(1)} = \frac{3}{25}. \end{align*}$

Compute the limit: $\lim _{x \to -2} \frac {\frac {3}{x+1} + \frac {6}{x+4}}{x+2}.$

When you plug in $x = -2$ , you get $\frac 00$

Add the fractions in the numerator

$\frac {a}{b} \pm \frac {c}{d} = \frac {ad \pm bc}{bd}$

To divide, multiply by the reciprocal: $\frac {a}{b} \div \frac {c}{d} = \frac {a}{b} \cdot \frac {d}{c}$

Cancel a common factor

The value of the limit is $\answer [given]{-\frac {9}{2}}$

Compute the limit: $\lim _{x \to 3} \frac {\frac {2}{x-5} + \frac {x+2}{5}}{x^2 - 5x + 6}.$

When you plug in $x = 3$ , you get $\frac 00$

Add the fractions in the numerator

$\frac {a}{b} \pm \frac {c}{d} = \frac {ad \pm bc}{bd}$

To divide, multiply by the reciprocal $\frac {a}{b} \div \frac {c}{d} = \frac {a}{b} \cdot \frac {d}{c}$

Cancel a common factor

The value of the limit is $\answer [given]{-\frac {3}{10}}$

Absolute Values

The definition of the absolute value is:

$\left | x \right | = \begin {cases} \hfill x & \text {if $x\geq 0$}\\ -x & \text {if $x<0$} \end {cases}$

To calculate a limit involving an absolute value, we will need to remove the absolute value bars. To do this correctly, we can see from the definition that it is necessary to know whether the quantity in the absolute value bars is positive or negative.

Compute the limit: $\displaystyle {\lim _{x \to 3^{-}} \frac {|x-3|}{x-3}}$ .
Plugging in $x = 3$ gives the indeterminate form $0/0$ . To resolve this limit, we will do a sign analysis on the quantity in the absolute value bars, $x-3$ .

Since $x \to 3^-$ , we have $x<3$ and hence $x-3 <0$ . Since the quantity in the absolute value bars is negative, we can compute its absolute value as follows:

$|x-3| = -(x-3).$

Using this in the limit, we get

$\begin{align*} \lim_{x \to 3^-} \frac{|x-3|}{x-3} &= \lim_{x \to 3^-} \frac{-(x-3)}{x-3} \\ &= \lim_{x \to 3^-} (-1) \\ &= -1. \end{align*}$

Compute the limit: $\lim _{x \to 4^-} \frac {4-x}{|x-4|}$

Since $x \to 4^-$ , we have $x<4$

Is $x-4$ positive or negative?

Remove the absolute value bars; include a negative sign if necessary

Simplify the fraction

The value of the limit is $\answer [given]{1}$

Compute the limit: $\lim _{x \to -3^-} \frac {|x+3|}{2x+6}$

Since $x \to -3^-$ , we have $x<-3$

Is $x+3$ positive or negative?

Remove the absolute value bars; include a negative sign if necessary

Simplify the fraction

The value of the limit is $\answer [given]{-1/2}$

Here are some detailed, lecture style videos on finding limits analytically: