Answer
See examples below.
Work Step by Step
Example 1.
$\displaystyle \lim_{x\rightarrow 1}\frac{1}{(x-1)}$ is a $\displaystyle \frac{k}{0}$ form, which is determinate,
but it doesn't exist, because
$\displaystyle \frac{k}{0^{+}}=\pm\infty, \qquad \frac{k}{Small}=Big $
the left sided limit is $-\infty$, and the right-sided one is $+\infty.$
Example 2.
$\displaystyle \lim_{x\rightarrow\infty}\frac{3x^{2}+x-85}{4x+1}$ is a $\displaystyle \frac{\infty}{\infty}$ form, which is indeterminate.
After applying theorem 10.2, we can discard all but the leading terms. Thus, we have the limit:
$=\displaystyle \lim_{x\rightarrow\infty}\frac{3x^{2}}{4x}=\lim_{x\rightarrow\infty}\frac{3x}{4}$$=+\infty$
