Answer
See below.
Work Step by Step
For example, $\displaystyle \lim_{x\rightarrow 1}\frac{(x-1)^{2}}{(x-1)}$ is a $\displaystyle \frac{0}{0}$ form.
After cancelling the common term, the limit equals
$=\displaystyle \lim_{x\rightarrow 1}(x-1)=0-0=0$
Another example, $\displaystyle \lim_{x\rightarrow\infty}\frac{3x^{2}+x-85}{4x^{2}+1}$ is a $\displaystyle \frac{\infty}{\infty}$ form.
After applying theorem 10.2, we can discard all but the leading terms and we get the limit:
$=\displaystyle \lim_{x\rightarrow\infty}\frac{3x^{2}}{4x^{2}}=\lim_{x\rightarrow\infty}\frac{3}{4}$=$\displaystyle \frac{3}{4}$
