Here we will look at specific examples using the limit.
Recall:
Suppose
\[
\lim_{x\to a}f(x)=L\,\,\text{ and }\,\,\lim_{x\to a}g(x)=M
\]
Then
\({\displaystyle \lim_{x\to a}\left[f(x)\right]}^{r}{\displaystyle =\left[\lim_{x\to a}f(x)\right]^{r}=L^{r}}\)
where \(r\) is a positive number.
\({\displaystyle \lim_{x\to a}cf(x)=c\lim_{x\to a}f(x)=cL}\) where
\(c\) is a real number.
\({\displaystyle \lim_{x\to a}\left[f(x)\pm g(x)\right]=\lim_{x\to a}f(x)\pm\lim_{x\to a}g(x)=L\pm M}\).
\({\displaystyle \lim_{x\to a}\left[f(x)g(x)\right]=\left[\lim_{x\to a}f(x)\right]\left[\lim_{x\to a}g(x)\right]=L\cdot M}\).
\({\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}=\frac{L}{M}}\)
where \(M\ne0\).
\({\displaystyle \lim_{x\to a}x}=a\)
\({\displaystyle \lim_{x\to a}b}=b\)
\({\displaystyle \lim_{x\to \infty}\frac{1}{x}}=0\) (thm. 5)
If \(f(x)=g(x)\) for all \(x\ne a\), then \({\displaystyle \lim_{x\to a}f(x)=\lim_{x\to a}g(x)}\). (cor. 1)
If \(\lim_{x\to a^{+}}f(x)\ne\lim_{x\to a^{-}}f(x)\), then we say \(\lim_{x\to a}f(x)\) does not exists. (cor. 2)
Example 1
Evaluate the limit:
\[
\lim_{x\to3}\dfrac{x^{2}+3x+4}{x^{2}+7}
\]
SOLUTION:
\[\begin{split}
\begin{align*}
\lim_{x\to3}\dfrac{x^{2}+3x+4}{x^{2}+7} & =\dfrac{\lim_{x\to3}x^{2}+\lim_{x\to3}3x+\lim_{x\to3}4}{\lim_{x\to3}x^{2}+\lim_{x\to3}7}\\
& =\dfrac{\left(\lim_{x\to3}x\right)^{2}+3\lim_{x\to3}x+4}{\left(\lim_{x\to3}x\right)^{2}+7}\\
& =\dfrac{\left(3\right)^{2}+3\cdot3+4}{\left(3\right)^{2}+7}\\
& =\frac{11}{8}
\end{align*}
\end{split}\]
Example 2
Evaluate the limit:
\[
\lim_{x\to64}\dfrac{\sqrt{x}-8}{x-64}
\]
SOLUTION:
\[\begin{split}
\begin{align*}
\lim_{x\to64}\dfrac{\sqrt{x}-8}{x-64} & =\lim_{x\to64}\dfrac{\sqrt{x}-8}{x-64}\cdot\left(\dfrac{\sqrt{x}+8}{\sqrt{x}+8}\right)\\
& =\lim_{x\to64}\dfrac{x-64}{(x-64)\cdot(\sqrt{x}+8)}\\
& =\lim_{x\to64}\dfrac{1}{\sqrt{x}+8}\\
& =\dfrac{\lim_{x\to64}1}{\lim_{x\to64}\sqrt{x}+\lim_{x\to64}8}\\
& =\dfrac{1}{\sqrt{\lim_{x\to64}x}+8}\\
& =\dfrac{1}{\sqrt{64}+8}\\
& =\dfrac{1}{8+8}\\
& =\dfrac{1}{16}
\end{align*}
\end{split}\]
Example 3
Let
\[\begin{split}
f(x)=\begin{cases}
x^{2}+1 & x\le1\\
3x-2 & x>1
\end{cases}
\end{split}\]
Evaluate:
\(f(1)\)
\({\displaystyle \lim_{x\to1}f(x)}\)
SOLUTION:
Evaluate \(f(1)\) means to evaluate \(f\) exactly when \(x=1\). According
to the definition of the piecewise function we have
\[
f(1)=1^{2}+1=2.
\]
In order to evaluate this piecewise function we must first evaluate
\(\lim_{x\to1^{-}}f(x)\) and \(\lim_{x\to1^{+}}f(x)\). If the two are
equal \(L\) then we say \(\lim_{x\to1}f(x)=L\); however, if the two
are not equal we say \(\lim_{x\to1}f(x)\) does not exist.
\[\begin{split}
\begin{aligned}\lim_{x\to1^{-}}f(x) & =\lim_{x\to1^{-}}\left(x^{2}+1\right)=1^{2}+1=2\\
\lim_{x\to1^{+}}f(x) & =\lim_{x\to1^{+}}\left(3x-2\right)=3(1)-2=1
\end{aligned}
\end{split}\]
Since \(\lim_{x\to1^{-}}f(x)\ne\lim_{x\to1^{+}}f(x)\) we say \(\lim_{x\to1}f(x)\)
does not exist.
Example 4
Evaluate the limit:
\[
\lim_{h\to0}\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}
\]
SOLUTION:
First, we will simplify \(\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}\)
\[\begin{split}
\begin{align*}
\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h} & =\dfrac{\left(\frac{x}{x}\right)\cdot\frac{1}{x+h}-\frac{1}{x}\cdot\left(\frac{x+h}{x+h}\right)}{h}\\
& =\dfrac{\frac{x-(x+h)}{x(x+h)}}{h}\\
& =\dfrac{\frac{-h}{x(x+h)}}{h}\\
& =\dfrac{1}{h}\cdot\left(\dfrac{-h}{x(x+h)}\right)\\
& =-\dfrac{1}{x(x+h)}
\end{align*}
\end{split}\]
Next, we notice that \(\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}=-\dfrac{1}{x(x+h)}\)
for all \(h\ne0\). Thus, by Corollary 1 we have
\[
\lim_{h\to0}\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h\to0}\dfrac{-1}{x(x+h)}=-\dfrac{1}{x(x+0)}=-\dfrac{1}{x^{2}}.
\]
Example 5
Evaluate the limit:
\[
\lim_{h\to0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h}
\]
SOLUTION:
First, we simplify \(\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\):
\[\begin{split}
\begin{align*}
\dfrac{\sqrt{x+h}-\sqrt{x}}{h} & =\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\\
& =\dfrac{(x+h)-(x)}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\\
& =\dfrac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\\
& =\dfrac{1}{\sqrt{x+h}+\sqrt{x}}
\end{align*}
\end{split}\]
Next, notice \(\dfrac{\sqrt{x+h}-\sqrt{x}}{h}=\dfrac{1}{\sqrt{x+h}+\sqrt{x}}\)
for all \(h\ne0\). Thus, by Corollary 1 we have
\[
\lim_{h\to0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h\to0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}}=\dfrac{1}{\sqrt{x+0}+\sqrt{x}}=\dfrac{1}{2\sqrt{x}}
\]