Question:
Evaluate
$\lim _{x \rightarrow \frac{1}{2}}\left(\frac{4 x^{2}-1}{2 x-1}\right)$
Solution:
To evaluate:
$\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}$
Formula used:
We have,
$\lim _{x \rightarrow a} f(x)=f(a)$
As $x \rightarrow \frac{1}{2}$, we have
$\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x+1)(2 x-1)}{2 x-1}$
$\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=\lim _{x \rightarrow \frac{1}{2}}(2 x+1)$
$\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=2$
Thus, the value of $\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}$ is 2 .