Question

the limit as x approaches 0 of (2x)/(sqrt x+1)-1) is 4 but how?

Answer 1

L'Hopital rull

Answer 2

\[\lim_{x \rightarrow 0}2*2\sqrt{x+1} = 4\]

Answer 3

\[\lim_{x \rightarrow 0} 2x \div (\sqrt{x+1}-1)\]
when you put X=0 then you will get 0/0 form.i.e this is indeterminate form..
Hence follow L_hospital rule(differentiate up and down with respect to variable X):
and you will get the value of limit is 4.

Answer 4

\[\frac{2x}{\sqrt{x+1}-1} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\frac{2x(\sqrt{x+1}+1)}{(x+1)-1}\]
\[\frac{2x(\sqrt{x+1}+1)}{x}=2(\sqrt{x+1}+1)\]

Answer 5

\[\lim_{x \rightarrow 0}\frac{2x}{\sqrt{x+1}-1}=\lim_{x \rightarrow 0}2(\sqrt{x+1}+1)=2(\sqrt{0+1}+1)\]