Find the value

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 .

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now