Solution:
Given:
[tex]\lim_{x\to0}(\frac{\sqrt{x+1}-1}{x})[/tex]
To solve, we have
step 1: From rationalization.
This gives:
[tex]\begin{gathered} \frac{1}{\sqrt{x+1}+1}\times\frac{\sqrt{x+1}-1}{\sqrt{x+1}-1} \\ =\frac{\sqrt{x+1}-1}{(\sqrt{x+1})^2-1} \\ =\frac{\sqrt{x+1}-1}{x+1-1} \\ \implies\frac{\sqrt{x+1}-1}{x} \\ This\text{ implies that} \\ \frac{\sqrt{x+1}-1}{x}\text{ is equaivalent to}\frac{1}{\sqrt{x+1}+1} \end{gathered}[/tex]
step 2: Substitute the value of zero for x.
Thus, we have
[tex]\begin{gathered} \lim_{x\to0}(\frac{1}{\sqrt{x+1}+1}) \\ when\text{ x =0, we have} \\ \frac{1}{\sqrt{0+1}+1}=\frac{1}{2} \end{gathered}[/tex]
Hence, we have
[tex]\lim_{x\to0}(\frac{\sqrt{x+1}-1}{x})=\frac{1}{2}[/tex]
