APPLY the L'Hopital rule. L = \(\displaystyle\lim_{x \rightarrow 0}\frac{ \dfrac{d}{DX}(\sqrt{1+x}-\sqrt{1-x})}{\frac{d}{dx}(x)}\)\(L =\frac{\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}{}}(-1)}{1}\)\(L={1\over 2 }+{1\over 2} = 1\)

"> APPLY the L'Hopital rule. L = \(\displaystyle\lim_{x \rightarrow 0}\frac{ \dfrac{d}{DX}(\sqrt{1+x}-\sqrt{1-x})}{\frac{d}{dx}(x)}\)\(L =\frac{\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}{}}(-1)}{1}\)\(L={1\over 2 }+{1\over 2} = 1\)

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\(\displaystyle\lim_{x \rightarrow 0} \dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}\) is given by

UPSEE Calculus in UPSEE 11 months ago

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Let L = \(\displaystyle\lim_{x \rightarrow 0} \dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}\)As it is \(\frac{0}{0}\) for APPLY the L'Hopital rule. L = \(\displaystyle\lim_{x \rightarrow 0}\frac{ \dfrac{d}{DX}(\sqrt{1+x}-\sqrt{1-x})}{\frac{d}{dx}(x)}\)\(L =\frac{\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}{}}(-1)}{1}\)\(L={1\over 2 }+{1\over 2} = 1\)

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Posted on 01 Dec 2024, this text provides information on UPSEE related to Calculus in UPSEE. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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