Answer
$$ \lim _{x \rightarrow \frac{\pi }{3}} \frac{2 \cos ^{2} x+3 \cos x-2}{2 \cos x-1} = \frac{5 }{2} $$
Work Step by Step
Given $$ \lim _{x \rightarrow \frac{\pi }{3}} \frac{2 \cos ^{2} x+3 \cos x-2}{2 \cos x-1}$$
let $$ f(x) = \frac{2 \cos ^{2} x+3 \cos x-2}{2 \cos x-1}$$
Since, we have
$$ f(\frac{\pi }{3})= \frac{2 \cos ^{2} \frac{\pi }{3}+3 \cos \frac{\pi }{3}-2}{2 \cos \frac{\pi }{3}-1}= \frac{ \frac{1}{2}+ \frac{3 }{2}-2}{1-1}=\frac{0}{0}$$
So, transform algebraically and cancel, we get
\begin{aligned}
L&=\lim _{x \rightarrow \frac{\pi }{3}} \frac{2 \cos ^{2} x+3 \cos x-2}{2 \cos x-1}\\
&= \lim _{x \rightarrow \frac{\pi }{3}} \frac{(2 \cos x-1) ( \cos x+2)}{2 \cos x-1}\\
&= \lim _{x \rightarrow \frac{\pi }{3}} ( \cos x+2)\\
&= ( \cos \frac{\pi }{3}+2) \\
&= \frac{1 }{2}+2 \\
&= \frac{5 }{2} \\
\end{aligned}
