Toggle navigation

.md .pdf

Contents

Limit Examples

Here we will look at specific examples using the limit.

Recall:

Suppose

$$\underset{x\to a}{lim}f(x)=L\text{ and }\underset{x\to a}{lim}g(x)=M$$

Then

1. ${{\displaystyle \underset{x\to a}{lim}[f(x)]}}^r{\displaystyle ={[\underset{x\to a}{lim}f(x)]}^r=L^r}$ where $r$ is a positive number.

2. ${\displaystyle \underset{x\to a}{lim}cf(x)=c\underset{x\to a}{lim}f(x)=cL}$ where $c$ is a real number.

3. ${\displaystyle \underset{x\to a}{lim}[f(x)\pm g(x)]=\underset{x\to a}{lim}f(x)\pm \underset{x\to a}{lim}g(x)=L\pm M}$.

4. ${\displaystyle \underset{x\to a}{lim}[f(x)g(x)]=[\underset{x\to a}{lim}f(x)][\underset{x\to a}{lim}g(x)]=L\cdot M}$.

5. ${\displaystyle \underset{x\to a}{lim}\frac{f(x)}{g(x)}=\frac{\underset{x\to a}{lim}f(x)}{\underset{x\to a}{lim}g(x)}=\frac{L}{M}}$ where $M\ne 0$.

6. ${\displaystyle \underset{x\to a}{lim}x}=a$

7. ${\displaystyle \underset{x\to a}{lim}b}=b$

8. ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}}=0$ (thm. 5)

9. If $f(x)=g(x)$ for all $x\ne a$, then ${\displaystyle \underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}g(x)}$. (cor. 1)

10. If $\underset{x\to a^{+}}{lim}f(x)\ne \underset{x\to a^{-}}{lim}f(x)$, then we say $\underset{x\to a}{lim}f(x)$ does not exists. (cor. 2)

---

Example 1

Evaluate the limit:

$$\underset{x\to 3}{lim}{\displaystyle \frac{x^2+3x+4}{x^2+7}}$$

SOLUTION:

$$\begin{array}{r}\begin{array}{rl}\underset{x\to 3}{lim}{\displaystyle \frac{x^2+3x+4}{x^2+7}}& ={\displaystyle \frac{\underset{x\to 3}{lim}{x}^2+\underset{x\to 3}{lim}3x+\underset{x\to 3}{lim}4}{\underset{x\to 3}{lim}{x}^2+\underset{x\to 3}{lim}7}}\\ & ={\displaystyle \frac{{\left(\underset{x\to 3}{lim}x\right)}^2+3\underset{x\to 3}{lim}x+4}{{\left(\underset{x\to 3}{lim}x\right)}^2+7}}\\ & ={\displaystyle \frac{{\left(3\right)}^2+3\cdot 3+4}{{\left(3\right)}^2+7}}\\ & =\frac{11}{8}\end{array}\end{array}$$

---

Example 2

Evaluate the limit:

$$\underset{x\to 64}{lim}{\displaystyle \frac{\sqrt{x}-8}{x-64}}$$

SOLUTION:

$$\begin{array}{r}\begin{array}{rl}\underset{x\to 64}{lim}{\displaystyle \frac{\sqrt{x}-8}{x-64}}& =\underset{x\to 64}{lim}{\displaystyle \frac{\sqrt{x}-8}{x-64}}\cdot \left({\displaystyle \frac{\sqrt{x}+8}{\sqrt{x}+8}}\right)\\ & =\underset{x\to 64}{lim}{\displaystyle \frac{x-64}{(x-64)\cdot (\sqrt{x}+8)}}\\ & =\underset{x\to 64}{lim}{\displaystyle \frac{1}{\sqrt{x}+8}}\\ & ={\displaystyle \frac{\underset{x\to 64}{lim}1}{\underset{x\to 64}{lim}\sqrt{x}+\underset{x\to 64}{lim}8}}\\ & ={\displaystyle \frac{1}{\sqrt{\underset{x\to 64}{lim}x}+8}}\\ & ={\displaystyle \frac{1}{\sqrt{64}+8}}\\ & ={\displaystyle \frac{1}{8+8}}\\ & ={\displaystyle \frac{1}{16}}\end{array}\end{array}$$

---

Example 3

Let

$$\begin{array}{r}f(x)=\{\begin{array}{ll}{x}^2+1& x\le 1\\ 3x-2& x>1\end{array}\end{array}$$

Evaluate:

1. $f(1)$

2. ${\displaystyle \underset{x\to 1}{lim}f(x)}$

SOLUTION:

1. Evaluate $f(1)$ means to evaluate $f$ exactly when $x=1$. According to the definition of the piecewise function we have

$$f(1)=1^2+1=2.$$

1. In order to evaluate this piecewise function we must first evaluate $\underset{x\to 1^{-}}{lim}f(x)$ and $\underset{x\to 1^{+}}{lim}f(x)$. If the two are equal $L$ then we say $\underset{x\to 1}{lim}f(x)=L$; however, if the two are not equal we say $\underset{x\to 1}{lim}f(x)$ does not exist.

$$\begin{array}{r}\begin{array}{rl}\underset{x\to 1^{-}}{lim}f(x)& =\underset{x\to 1^{-}}{lim}(x^2+1)=1^2+1=2\\ \underset{x\to 1^{+}}{lim}f(x)& =\underset{x\to 1^{+}}{lim}(3x-2)=3(1)-2=1\end{array}\end{array}$$

Since $\underset{x\to 1^{-}}{lim}f(x)\ne \underset{x\to 1^{+}}{lim}f(x)$ we say $\underset{x\to 1}{lim}f(x)$ does not exist.

---

Example 4

Evaluate the limit:

$$\underset{h\to 0}{lim}{\displaystyle \frac{\frac{1}{x+h}-\frac{1}{x}}{h}}$$

SOLUTION:

First, we will simplify ${\displaystyle \frac{\frac{1}{x+h}-\frac{1}{x}}{h}}$

$$\begin{array}{r}\begin{array}{rl}{\displaystyle \frac{\frac{1}{x+h}-\frac{1}{x}}{h}}& ={\displaystyle \frac{\left(\frac{x}{x}\right)\cdot \frac{1}{x+h}-\frac{1}{x}\cdot \left(\frac{x+h}{x+h}\right)}{h}}\\ & ={\displaystyle \frac{\frac{x-(x+h)}{x(x+h)}}{h}}\\ & ={\displaystyle \frac{\frac{-h}{x(x+h)}}{h}}\\ & ={\displaystyle \frac{1}{h}}\cdot \left({\displaystyle \frac{-h}{x(x+h)}}\right)\\ & =-{\displaystyle \frac{1}{x(x+h)}}\end{array}\end{array}$$

Next, we notice that ${\displaystyle \frac{\frac{1}{x+h}-\frac{1}{x}}{h}}=-{\displaystyle \frac{1}{x(x+h)}}$ for all $h\ne 0$. Thus, by Corollary 1 we have

$$\underset{h\to 0}{lim}{\displaystyle \frac{\frac{1}{x+h}-\frac{1}{x}}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{-1}{x(x+h)}}=-{\displaystyle \frac{1}{x(x+0)}}=-{\displaystyle \frac{1}{x^2}}.$$

---

Example 5

Evaluate the limit:

$$\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{h}}$$

SOLUTION:

First, we simplify ${\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{h}}$:

$$\begin{array}{r}\begin{array}{rl}{\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{h}}& ={\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{h}}\cdot \left({\displaystyle \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}}\right)\\ & ={\displaystyle \frac{(x+h)-(x)}{h(\sqrt{x+h}+\sqrt{x})}}\\ & ={\displaystyle \frac{h}{h(\sqrt{x+h}+\sqrt{x})}}\\ & ={\displaystyle \frac{1}{\sqrt{x+h}+\sqrt{x}}}\end{array}\end{array}$$

Next, notice ${\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{h}}={\displaystyle \frac{1}{\sqrt{x+h}+\sqrt{x}}}$ for all $h\ne 0$. Thus, by Corollary 1 we have

$$\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{1}{\sqrt{x+h}+\sqrt{x}}}={\displaystyle \frac{1}{\sqrt{x+0}+\sqrt{x}}}={\displaystyle \frac{1}{2\sqrt{x}}}$$

previous

Limits

next

Limits and Continuity