Calculus - Limits

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$\hspace{-20}\bf{(A)::\; }$Given $\bf{\lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e}{x}}$\\\\\\ Now we can write $\bf{(1+x)^{\frac{1}{x}}=e^{\log_{e}(1+x)^{\frac{1}{x}}}=e^{\frac{1}{x}\cdot \log_{e}(1+x)}}$\\\\\\ Now Using $\bf{\bullet \; \log_{e}(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}+.........\infty}$\\\\\\ So $\bf{\lim_{x\rightarrow 0}\frac{\log_{e}(1+x)}{x}=\lim_{x\rightarrow 0}1-\frac{x}{2}+\frac{x^2}{3}+.....\infty \approx 1-\frac{x}{2}}$\\\\\\ So $\bf{(1+x)^{\frac{1}{x}}=e^{\frac{1}{x}\cdot \log_{e}(1+x)}=e^{1-\frac{x}{2}}=e^{1}\cdot e^{-\frac{x}{2}}}$\\\\\\ So $\bf{\lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e}{x}=\lim_{x\rightarrow 0}\frac{e^{1}\cdot e^{-\frac{x}{2}}-e}{x}}$\\\\\\ $\bf{=\lim_{x\rightarrow 0}e\cdot \left(\frac{e^{-\frac{x}{2}}-1}{x}\right)= \lim_{x\rightarrow 0} e\cdot \left(\frac{e^{-\frac{x}{2}}-1}{\frac{-x}{2}}\right)\cdot -\frac{1}{2}=-\frac{e}{2}}$\\\\\\ bcz $\bf{\lim_{y\rightarrow 0}\frac{e^y-1}{y}=1.}$$

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Raunaq Saha but how is lim 1-x 2+x2 3+...âˆž â‰ˆ1-x 2 xâ†’0 shouldn't it be 1 ?
Upvote·0· Reply ·May 27 at 8:43
Raunaq Saha but how is lim 1-x/2+....âˆž â‰ˆ1-x/2 xâ†’0 shouldn't it be 1 ?
Upvote·0· Reply ·May 27 at 8:45
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