Question:
$\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}$
Solution:
Given $\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}$
The above equation can be written as
$=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x)^{2}-1}{2 x-1}$
Using $a^{2}-b^{2}$ formula and expanding we get
$=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)(2 x+1)}{2 x-1}$
On simplifying and applying the limits we get
$\Rightarrow \lim _{x \rightarrow \frac{1}{2}}(2 x+1)=2$
$\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=2$
On simplifying and applying the limits we get
$\Rightarrow \lim _{x \rightarrow \frac{1}{2}}(2 x+1)=2$
$\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=2$