Question

$\lim_{x\to \ 4} \frac{x-4}{\sqrt{x}-\sqrt{4} }$ Please answer this one

Answer 1

Answer:

$\large \boxed{\sf \ \ \lim_{x\to \ 4} \dfrac{x-4}{\sqrt{x}-\sqrt{4} }=4 \ \ }$

Step-by-step explanation:

Hello,

We need to find the following limit.

$\displaystyle \lim_{x\to \ 4} \dfrac{x-4}{\sqrt{x}-\sqrt{4} }$

First of all, for any x real number different from 4 and positive, we can write

$\dfrac{x-4}{\sqrt{x}-\sqrt{4}} = \dfrac{(x-4)(\sqrt{x}+\sqrt{4})} {(\sqrt{x}-\sqrt{4})(\sqrt{x}+\sqrt{4})}} ==\dfrac{(x-4)(\sqrt{x}+\sqrt{4})}{x-4}=\sqrt{x}+\sqrt{4}$

so the limit is

$\sqrt{4}+\sqrt{4}=2+2=4$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you