Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$
✓
Answer
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}}-1\big)}{\text{x}}\times\frac{\big(\sqrt{1+\text{x}}+1\big)}{\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x}-1)}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{1+\text{x}}+1\big)}$
$=\frac{1}{\sqrt{1}+1}=\frac12$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}}-1\big)}{\text{x}}\times\frac{\big(\sqrt{1+\text{x}}+1\big)}{\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x}-1)}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{1+\text{x}}+1\big)}$
$=\frac{1}{\sqrt{1}+1}=\frac12$
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