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Prove the following

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

$\Rightarrow \lim _{x \rightarrow \frac{1}{2}}(2 x+1)=2$

$\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=2$

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