Question

Evaluate
$\rm \displaystyle \lim_{n\to 1/4}\ \frac{4x-1}{2\sqrt{x}-1}$

Answer 1

See the image and ask if there is any problem

Answer 2

Answer:

$\displaystyle\lim_{x\to 1/4} \frac{4x-1}{2\sqrt{x}-1}=2$

Step-by-step explanation:

$\displaystyle \lim_{x\to 1/4} \frac{4x-1}{2\sqrt{x}-1}$

$=\displaystyle \lim_{x\to 1/4} \frac{4(x-\frac{1}{4})}{2(\sqrt{x}-\frac{1}{2})}$

$=\displaystyle \lim_{x\to 1/4} \frac{2(x-\frac{1}{4})}{\sqrt{x}-\sqrt{\frac{1}{4}}}$

$=\displaystyle \lim_{x\to 1/4}\ \frac{2}{\frac{\sqrt{x}-\sqrt{1/4}}{x-\frac{1}{4}}}$

$=\displaystyle\frac{\lim_{x\to 1/4}\:2}{\lim_{x\to 1/4}\:\frac{\sqrt{x}-\sqrt{\frac{1}{4}}}{x-\frac{1}{4}}}$

$=\displaystyle\frac{2}{\lim_{x\to 1/4}\:\frac{\sqrt{x}-\sqrt{\frac{1}{4}}}{x-\frac{1}{4}}}$

Using

$\boxed{\bf\lim_{x\to\,a}\:\frac{x^n-a^n}{x-a}=na^{n-1}}$

$=\displaystyle\frac{2}{\frac{1}{2}(\frac{1}{4})^{\frac{1}{2}-1}}$

$=\displaystyle\frac{2}{\frac{1}{2}(\frac{1}{4})^{\frac{-1}{2}}}$

$=\displaystyle\frac{2}{\frac{1}{2}(4)^{\frac{1}{2}}}$

$=\displaystyle\frac{2}{\frac{1}{2}(2)}$

$=\displaystyle\frac{2}{1}$

$=2$

$\implies\boxed{\lim_{x\to 1/4} \frac{4x-1}{2\sqrt{x}-1}=2}$