What is $${\lim_{x\rightarrow 0}} \frac{(1+x)^{1/x}-e+\frac{ex}2}{x^2}$$ Here's what I tried $$\text {ATTEMPT}$$ We know that ${\lim_{h\rightarrow 0}} (1+h)^{1/h}=e$ so the $e.-e$ get cancelled out now we want ${\lim_{x\rightarrow 0}} \frac{ex/2}{x^2}$ which is $\infty$ but it isn't in any option given. So please tell me any approximation or any better way to evaluate it.
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What does it mean $h$? – openspace Apr 10 '16 at 11:36
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Sorry its x ill edit – Archis Welankar Apr 10 '16 at 11:46
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2By imposing that the solution should not use l'Hospital's theorem you make the problem much more difficult than it really is. – Alex M. Apr 10 '16 at 12:03
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I agree: the numerator is a Taylor expansion, so that algebraic tricks or fundamental limits seem to be insufficient. – Siminore Apr 10 '16 at 12:10
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It is in my book in sidenote – Archis Welankar Apr 10 '16 at 12:10
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@Archis Should ${\lim_{\color{red}{x}\rightarrow 0}} (1+h)^{1/h}=e$ be ${\lim_{\color{red}{h}\rightarrow 0}} (1+h)^{1/h}=e$? – BLAZE Apr 10 '16 at 12:40
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Possible duplicate of The limit of $((1+x)^{1/x} - e+ ex/2)/x^2$ as $x\to 0$ – Arnaud D. Dec 04 '19 at 17:12
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$$(1+x)^{\frac{1}{x}}=e^{\ln((1+x)^{\frac{1}{x}})}=e^{1-\frac{x}{2}+\frac{x^2}{3}...}$$ $$=e.e^{-\frac{x}{2}+\frac{x^2}{3}...}$$ $$=e\Big[1+\Big(-\frac{x}{2}+\frac{x^2}{3}...\Big)+\Big(-\frac{x}{2}+\frac{x^2}{3}...\Big)^2 +...\Big]$$ $$=e\Big[1-\frac{x}{2}+\frac{11x^2}{24}...\Big]$$
$$\therefore \lim_{x\to 0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}=\frac{11e}{24}$$
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Nice method (+1) but you could improve it a little by showing the intermediate steps of how you got to $\lim_{x\to 0}\dfrac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}=\dfrac{11e}{24}$. As it is a pretty big jump. – BLAZE Apr 10 '16 at 13:00