## Section3.4Computing Limits: Algebraically

### Subsection3.4.1Properties of limits

We begin by deriving a handful of theorems to give us the tools to compute many limits without explicitly working with the precise definition of a limit.

Roughly speaking, these rules say that to compute the limit of an algebraic expression, it is enough to compute the limits of the “innermost bits” and then combine these limits. This often means that it is possible to simply plug in a value for the variable, since $\ds \lim_{x\to a} x =a\text{.}$

###### Example3.10. Limit Properties.

Compute $\ds\lim_{x\to 1}{x^2-3x+5\over x-2}\text{.}$

Solution

If we apply the theorem in all its gory detail, we get

\begin{equation*} \begin{split} \lim_{x\to 1}{x^2-3x+5\over x-2} = \amp{\ds\lim_{x\to 1}(x^2-3x+5)\over \ds\lim_{x\to1}(x-2)}\\ =\amp {(\ds\lim_{x\to 1}x^2)-(\ds\lim_{x\to1}3x)+(\ds\lim_{x\to1}5)\over (\ds\lim_{x\to1}x)-(\ds\lim_{x\to1}2)}\\ =\amp {(\ds\lim_{x\to 1}x)^2-3(\ds\lim_{x\to1}x)+5\over (\ds\lim_{x\to1}x)-2}\\ =\amp {1^2-3\cdot1+5\over 1-2}\\ =\amp {1-3+5\over -1} = -3 \end{split} \end{equation*}

It is worth commenting on the trivial limit $\ds \lim\limits_{x\to1}5\text{.}$ From one point of view this might seem meaningless, as the number 5 can't “approach” any value, since it is simply a fixed number. However, 5 can, and should, be interpreted here as the function that has value 5 everywhere, $f(x)=5\text{,}$ with graph a horizontal line. From this point of view it makes sense to ask what happens to the values of the function (height of the graph) as $x$ approaches 1.

We're primarily interested in limits that aren't so easy, namely, limits in which a denominator approaches zero. There are a handful of algebraic tricks that work on many of these limits.

We first look at two cases when the limit of the quotient $\dfrac{f(x)}{g(x)}$ as $x$ approaches $a$ takes on the indeterminate form $\dfrac{0}{0}\text{.}$

###### Example3.11. Zero Denominator.

Compute $\ds\lim_{x\to1}{x^2+2x-3\over x-1}\text{.}$

Solution

We can't simply plug in $x=1$ because that makes the denominator zero. However:

\begin{equation*} \begin{split} \lim_{x\to1}{x^2+2x-3\over x-1} =\amp \lim_{x\to1}{(x-1)(x+3)\over x-1}\\ =\amp \lim_{x\to1}(x+3)=4 \end{split} \end{equation*}

The technique used to solve the previous example can be referred to as factor and cancel. Its validity comes from the fact that we are allowed to cancel $x-1$ from the numerator and denominator. Remember in Calculus that we have to make sure we don't cancel zeros, so we require $x-1\neq 0$ in order to cancel it. But looking back at the definition of a limit using $x\to 1\text{,}$ the key point for this example is that we are taking values of $x$ close to $1$ but not equal to $1\text{.}$ This is exactly what we wanted ($x\neq 1$) in order to cancel this common factor.

###### Example3.12. Another Zero Denominator.

Compute

\begin{equation*} \lim_{x \to 0} \dfrac{x}{x^{3}} \end{equation*}
Solution

We can't simply plug in $x = 0$ because that makes the denominator zero. Let's simplify algebraically first to get

\begin{equation*} \lim_{x \to 0} \dfrac{x}{x^{3}} = \lim_{x \to 0} \dfrac{1}{x^{2}}\text{.} \end{equation*}

As $x$ approaches zero from either the left or right side, the value of $x^{2}$ approaches zero and so $\dfrac{1}{x^{2}}$ approaches infinity. In other words, $\lim_{x \to 0} \dfrac{x}{x^{3}}$ does not exist since it does not approach a real number.

The last two examples show that limits of indeterminate form $\dfrac{0}{0}$ either exist or do not, which is why this type of limit is called indeterminate.

While Theorem 3.9 is very helpful, we need a bit more to work easily with limits. Since the theorem applies when some limits are already known, we need to know the behavior of some functions that cannot themselves be constructed from the simple arithmetic operations of the theorem, such as $\ds\sqrt{x}\text{.}$ Also, there is one other extraordinarily useful way to put functions together: composition. If $f(x)$ and $g(x)$ are functions, we can form two functions by composition: $f(g(x))$ and $g(f(x))\text{.}$ For example, if $\ds f(x)=\sqrt{x}$ and $\ds \ds g(x)=x^2+5\text{,}$ then $\ds f(g(x))=\sqrt{x^2+5}$ and $\ds g(f(x))=(\sqrt{x})^2+5=x+5\text{.}$ Here is a companion to Theorem 3.9 for composition:

Note the special form of the condition on $f\text{:}$ it is not enough to know that $\ds\lim_{x\to L}f(x) = M\text{,}$ though it is a bit tricky to see why. We have included an example in the exercise section to illustrate this tricky point for those who are interested. Many of the most familiar functions do have this property, and this theorem can therefore be applied. For example:

This theorem is not too difficult to prove from the definition of limit.

Another of the most common algebraic tricks is called rationalization. Rationalizing makes use of the difference of squares formula $(a-b)(a+b)=a^2-b^2\text{.}$ Here is an example.

###### Example3.15. Rationalizing.

Compute $\ds\lim_{x\to-1} {\sqrt{x+5}-2\over x+1}\text{.}$

Solution
\begin{equation*} \begin{split} \lim_{x\to-1} {\sqrt{x+5}-2\over x+1} =\amp \lim_{x\to-1} {\sqrt{x+5}-2\over x+1}\cdot{\sqrt{x+5}+2\over \sqrt{x+5}+2}\\ =\amp \lim_{x\to-1} {x+5-4\over (x+1)(\sqrt{x+5}+2)}\\ =\amp \lim_{x\to-1} {x+1\over (x+1)(\sqrt{x+5}+2)}\\ =\amp \lim_{x\to-1} {1\over \sqrt{x+5}+2}={1\over4} \end{split} \end{equation*}

At the very last step we have used Theorems 3.13 and 3.14.

###### Example3.16. Left and Right Limit.

Evaluate $\ds\lim_{x\to 0^+}{x\over|x|}\text{.}$

Solution

The function $f(x)=x/|x|$ is undefined at 0; when $x>0\text{,}$ $|x|=x$ and so $f(x)=1\text{;}$ when $x\lt 0\text{,}$ $|x|=-x$ and $f(x)=-1\text{.}$ Thus

\begin{equation*} \ds \lim_{x\to 0^-}{x\over|x|}=\lim_{x\to 0^-}-1=-1 \end{equation*}

while

\begin{equation*} \ds \lim_{x\to 0^+}{x\over|x|}=\lim_{x\to 0^+}1=1\text{.} \end{equation*}

The limit of $f(x)$ must be equal to both the left and right limits; since they are different, the limit $\ds \lim_{x\to 0}{x\over|x|}$ does not exist.

###### Example3.17. Limit of Piecewise-defined Function.

Let

\begin{equation*} g(x) = \begin{cases} x + 3 \amp \text{ if } x \neq 1 \\ 2 \amp \text{ if } x = 1 \end{cases} \end{equation*}

Evaluate $\lim\limits_{x \to 1} g(x)\text{.}$

Solution

The domain of $g$ is the set of all real numbers. From the graph shown below, we see that $g(x)$ can be made as close to $4$ as we please by taking $x$ sufficiently close to $1\text{.}$ Therefore,

\begin{equation*} \lim_{x \to 1} g(x) = \lim_{x \to 1} (x+3) = 4 \end{equation*}

Observe that $g(1) = 2\text{,}$ which does not equal the value of the limit of the function $g$ as $x$ approaches $1\text{.}$ The value of $g(x)$ at $x = 1$ has no bearing on the existence or value of the limit as $g$ approaches $1\text{.}$

##### Exercises for Section 3.4.

Compute the limits. If a limit does not exist, explain why.

1. $\ds \lim_{x\to 3}{x^2+x-12\over x-3}$

$\lim\limits_{x \to 3} \dfrac{x^{2} + x - 12}{x - 3} = \lim\limits_{x \to 3} \dfrac{(x-3)(x+4)}{x-3} = \lim\limits_{x \to 3} x + 4 = 7\text{.}$
2. $\ds \lim_{x\to 1}{x^2+x-12\over x-3}$

$\lim\limits_{x \to 1} \dfrac{x^{2} + x - 12}{x - 3} = \lim\limits_{x \to 1} x + 4 = 5\text{.}$
3. $\ds \lim_{x\to -4}{x^2+x-12\over x-3}$

$\lim\limits_{x \to -4} \dfrac{x^{2} + x - 12}{x - 3} = \lim\limits_{x \to -4} x + 4 = 0\text{.}$
4. $\ds \lim_{x\to 2} {x^2+x-12\over x-2}$

$\lim\limits_{x \to 2^{+}} \dfrac{x^{2} + x - 12}{x - 2} = \dfrac{-6}{+0} = - \infty\text{.}$ $\lim\limits_{x \to 2^{-}} \dfrac{x^{2} + x - 12}{x - 2} = \dfrac{-6}{-0} = + \infty\text{.}$ Therefore, $\lim\limits_{x \to 2} \dfrac{x^{2} + x - 12}{x - 2}$ DNE.

5. $\ds \lim_{x\to 1} {\sqrt{x+8}-3\over x-1}$

$\lim\limits_{x\to 1} \frac{\sqrt{x+8}-3}{x-1} = \lim\limits_{x\to 1} \frac{\sqrt{x+8}-3}{x-1} \cdot \frac{\sqrt{x+8}+3}{\sqrt{x+8}+3} = \lim\limits_{x\to 1} \frac{(x-1)}{(x-1)(\sqrt{x+8}_3)} = \frac{1}{6}$
6. $\ds \lim_{x\to 0^+} \sqrt{{1\over x}+2} - \sqrt{1\over x}$

$\lim\limits_{x\to 0^+} \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}} = \lim\limits_{x\to 0^+} \frac{2}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}} = \lim\limits_{x\to 0^+} \frac{2\sqrt{x}}{\sqrt{1+2x}+1} = \frac{0}{2} = 0$
7. $\ds\lim _{x\to 2} 3$

$\lim\limits_{x\to 2} 3 = 3$
8. $\ds\lim _{x\to 4 } 3x^3 - 5x$

$\lim\limits_{x\to 4}(3x^3-5x) = 3(4^3)-5(4) = 172$
9. $\ds \lim _{x\to 0 } {4x - 5x^2\over x-1}$

$\lim\limits_{x\to 0} \frac{4x-5x^2}{x-1} = \frac{0}{-1} = 0$
10. $\ds\lim _{x\to 1 } {x^2 -1 \over x-1 }$

$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1} = \lim\limits_{x\to 1}\dfrac{(x-1)(x+1)}{x-1} = \lim\limits_{x\to 1} (x+1) = 2$
11. $\ds\lim _{x\to 0^ + } {\sqrt{2-x^2 }\over x}$

$\lim\limits_{x\to 0^+} \dfrac{\sqrt{2-x^2}}{x} = \lim\limits_{x\to 0^+}\dfrac{\frac{\sqrt{2-x^2}}{x}}{\frac{x}{x}} = \lim\limits_{x\to 0^+} \dfrac{\sqrt{\frac{2-x^2}{x^2}}}{\frac{x}{x}} = \lim\limits_{x\to 0^+}\sqrt{\frac{2}{x^2}-1} = \infty\text{.}$
12. $\ds\lim _{x\to 0^ + } {\sqrt{2-x^2}\over x+1}$

$\lim\limits_{x\to 0^+} \frac{\sqrt{2-x^2}}{x+1} = \frac{\sqrt{2-0}}{1} = \sqrt{2}$
13. $\ds\lim _{x\to a } {x^3 -a^3\over x-a}$

$\lim\limits_{x\to a} \dfrac{x^3-a^3}{x-a} = \lim\limits_{x \to a} \dfrac{(x-a)(a^2+ax+x^2)}{x-a} = 3a^2$
14. $\ds\lim _{x\to 2 } (x^2 +4)^3$

$\lim\limits_{x\to 2} (x^2+1)^3 = 5^3$

Let $f(x)=\left\{ \begin{array}{cc} 1 \amp \text{ if } x\neq 0 \\ 0 \amp \text{ if } x=0 \end{array} \right.$ and $g(x)=0\text{.}$ What are the values of $L=\lim\limits_{x\to 0}g(x)$ and $M=\lim\limits_{x\to L}f(x)\text{?}$ Is it true that $\lim\limits_{x\to 0}f(g(x))=M\text{?}$ What are some noteworthy differences between this example and Theorem 3.13?

$L=0$ and $M=1\text{.}$ No.
Solution

Graphically, we see that $g(x)$ is the line which lies along the $x$-axis, and $f(x)$ is the piecewise-defined function shown below.

Therefore, we find:

\begin{equation*} L = \lim_{x\to 0} g(x) = 0\text{,} \end{equation*}

and so

\begin{equation*} M = \lim_{x\to L} f(x) = \lim_{x\to 0} f(x) = 1\text{.} \end{equation*}

But,

\begin{equation*} \lim_{x\to 0} f(g(x)) = \lim_{x\to 0} f(0) = 0\text{.} \end{equation*}

The difference between this example and Theorem 3.13 is that we do not have $\lim\limits_{x\to 0} f(x) = f(0)\text{.}$

Sketch the graph of the given function $f$ and evaluate $\lim\limits_{x \to a} f(x)\text{,}$ if it exists, for the given values of $a\text{.}$

1. \begin{equation*} f(x) = \begin{cases} x + 1 \amp \text{ if } x \leq 0 \\ 1 \amp \text{ if } x > 0 \end{cases} ,\ a = 0 \end{equation*}
\begin{equation*} \begin{split} \amp \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} (x + 1) = 1 \amp \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} (1) = 1 \end{split} \end{equation*}

Therefore $\lim\limits_{x \to 0} f(x)= 1\text{.}$

2. \begin{equation*} f(x) = \begin{cases} x \amp \text{ if } x \lt 1 \\ -1 \amp \text{ if } x = 1 \\ -x + 2 \amp \text{ if } x > 1 \end{cases} , \ a = 1 \end{equation*}
\begin{equation*} \begin{split} \amp \lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} (-x+2) = 1 \amp \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} x = 1 \end{split} \end{equation*}

Therefore $\lim\limits_{x \to 1} f(x)= 1\text{.}$

3. \begin{equation*} f(x) = \begin{cases} \lvert x \lvert \amp \text{ if } x \neq 0 \\ 2 \amp \text{ if } x = 0 \end{cases} ,\ a = 0 \end{equation*}
\begin{equation*} \begin{split} \amp \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} x = 0 \amp \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} (-x) = 0 \end{split} \end{equation*}

Therefore $\lim\limits_{x \to 0} f(x)= 0\text{.}$

Find the indicated limit given that

\begin{equation*} \lim_{x \to a} p(x) = 3 \ \ \text{ and } \ \ \lim_{x \to a} q(x) = 4\text{.} \end{equation*}
1. $\lim\limits_{x \to a} \left[p(x) - q(x) \right]$

$\lim\limits_{x \to a} [ p(x) -q(x)] = \lim\limits_{x \to a} p(x) - \lim\limits_{x \to a} q(x) = 3-4=-1$
2. $\lim\limits_{x \to a} \sqrt{q(x)}$

$\lim\limits_{x \to a} \sqrt{q(x)} = \sqrt{\lim\limits_{x \to a} q(x)} = \sqrt{4} = 2$
3. $\lim\limits_{x \to a} \left[\dfrac{2p(x) - q(x)}{p(x)q(x)} \right]$

$\lim\limits_{x \to a} \left[ \dfrac{2p(x)-q(x)}{p(x)q(x)} \right] = \dfrac{2\lim\limits_{x \to a} p(x) - \lim\limits_{x \to a} q(x)}{ \lim\limits_{x \to a} p(x) \cdot \lim\limits_{x \to a} q(x)} = \dfrac{2(3)-4}{3 \cdot 4} = \dfrac{2}{12} = \dfrac{1}{6}$

Find the indicated limit, if it exists.

1. $\lim\limits_{x \to 0} \dfrac{x^{3}-x^2}{x^2}$

$\lim\limits_{x\to 0} \dfrac{x^3-x^2}{x^2} = \lim\limits_{x\to 0} (x-1) = -1$
2. $\lim\limits_{x \to 1} \dfrac{x^2}{x(x-1)}$

$\lim\limits_{x\to 1} \dfrac{x^2}{x(x-1)} = \lim\limits_{x\to 1} \dfrac{x}{(x-1)} = \lim\limits_{x\to 1} \dfrac{1}{1-\frac{1}{x}}$ DNE
3. $\lim\limits_{x \to 1} \dfrac{x-1}{\sqrt{x}-1}$

$\lim\limits_{x\to 1} \dfrac{x-1}{\sqrt{x}-1} = \lim\limits_{x\to 1} \dfrac{x-1}{\sqrt{x}-1} \cdot \dfrac{\sqrt{x}+1}{\sqrt{x}+1} = \lim\limits_{x\to 1}\dfrac{(x-1)(\sqrt{x}+1)}{x-1} = \lim\limits_{x\to 1}\left(\sqrt{x}+1\right) = 2$
4. $\lim\limits_{x \to -1} \dfrac{x+1}{x^{3}+1}$

$\lim\limits_{x\to -1} \dfrac{x+1}{x^3+1} = \lim\limits_{x\to -1} \dfrac{x+1}{(x+1)(x^2-x+1)} = \lim\limits_{x\to -1} \dfrac{1}{x^2-x+1} = \dfrac{1}{3}$
5. $\lim\limits_{x \to 1} \dfrac{x^{3}-3x+2}{x^{3}-x^{2}-x+1}$

$\lim\limits_{x\to 1} \dfrac{x^3-3x+2}{x^3-x^2-x+1} = \lim\limits_{x\to 1}\dfrac{(x-1)^2(x+2)}{(x-1)^2(x+1)} = \lim\limits_{x\to 1} \dfrac{x+2}{x+1} = \dfrac{3}{2}$
6. $\lim\limits_{x \to 0} \dfrac{x}{\sqrt{x+1}-1}$

\begin{equation*} \begin{split} \lim\limits_{x\to 0} \dfrac{x}{\sqrt{x+1}-1} \amp= \lim\limits_{x\to 0} \dfrac{x}{\sqrt{x+1}-1} \cdot \dfrac{\sqrt{x+1}+1}{\sqrt{x+1}+1} \\ \amp= \lim\limits_{x\to 0} \dfrac{x(\sqrt{x+1}+1)}{x} = \lim\limits_{x\to 0} \left(\sqrt{x+1}+1\right) = 2\end{split} \end{equation*}

Find the indicated one-sided limit, if it exists.

1. $\lim\limits_{x \to 1^{-}} \left(2x + 4\right)$

$\lim\limits_{x\to 1^-} (2x+4) = 2(1)+4 = 6$
2. $\lim\limits_{x \to 2^{+}} \dfrac{x-3}{x+2}$

$\lim\limits_{x\to 2^+} \dfrac{x-3}{x+2} = \dfrac{(2-3)}{(2+2)} = \dfrac{-1}{4}$
3. $\lim\limits_{x \to 0^{-}} \dfrac{1}{x}$

$\lim\limits_{x\to 0^-} \dfrac{1}{x} = -\infty\text{.}$ We refer to the graph of $\dfrac{1}{x}\text{:}$
4. $\lim\limits_{x \to 0^{-}} \dfrac{x-1}{x^{2}+1}$

$\lim\limits_{x\to 0^-} \dfrac{x-1}{x^2+1} = \dfrac{0-1}{0+1} = -1$
5. $\lim\limits_{x \to 1^{+}} \dfrac{1+x}{1-x}$

Since $\lim\limits_{x\to 1^+} (1+x) = 2$ and $\lim\limits_{x\to 1^+} \dfrac{1}{(1-x)} = -\infty\text{,}$ we have $\lim\limits_{x\to 1^+} \dfrac{1+x}{1-x} = -\infty$
6. $\lim\limits_{x \to 0^{+}} f(x)$ and $\lim\limits_{x \to 0^{-}} f(x)$ where
$\lim\limits_{x\to 0^+} f(x) = \lim\limits_{x\to 0^+} 2x = 0\text{,}$ and $\lim\limits_{x\to 0^-} f(x) = \lim\limits_{x\to 0^-} x^2 = 0$
7. $\lim\limits_{x \to 0^{+}} f(x)$ and $\lim\limits_{x \to 0^{-}} f(x)$ where
$\lim\limits_{x\to 0^+} f(x) = \lim\limits_{x\to 0^+} (-x+1)=1$$\lim\limits_{x\to 0^-} f(x) = \lim\limits_{x\to 0^-} (2x+3)=3$