Question
Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{\sqrt{1+\text{x}^{3}}-\sqrt{1-\text{x}^{3}}}{\text{x}^{2}}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{\sqrt{1+\text{x}^{3}}-\sqrt{1-\text{x}^{3}}}{\text{x}^{2}}$
✓
Answer
Given that $\lim\limits_{\text{x} \rightarrow 0}\frac{\sqrt{1+\text{x}^{3}}-\sqrt{1-\text{x}^{3}}}{\text{x}^{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big[\sqrt{1+\text{x}^{3}}-\sqrt{1-\text{x}^{3}}\Big]\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-3^{3}}\Big]}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]} $
$=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x}^{3})-(1-\text{x}^{3})}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{1+\text{x}^{3}-1-\text{x}^{3}}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]}$
$=\lim\limits_{\text{x} \rightarrow 1}\text{f}(\text{x})$
Hence, the required answer is 0.
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big[\sqrt{1+\text{x}^{3}}-\sqrt{1-\text{x}^{3}}\Big]\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-3^{3}}\Big]}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]} $
$=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x}^{3})-(1-\text{x}^{3})}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{1+\text{x}^{3}-1-\text{x}^{3}}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]}$
$=\lim\limits_{\text{x} \rightarrow 1}\text{f}(\text{x})$
Hence, the required answer is 0.
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