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How do I solve the following limit.

$$ \lim_{x\rightarrow 0}\frac{e - (1+x)^{\frac{1}{x}}}{x}$$

The answer of the limit should be $\frac{e}{2}$.The problem is not that I’m unable to solve but the problem is that I’m even unable to think how to proceed in the first place. I am having no clue how to approach this and other problem like this.

Along with providing the solution for this problem can you please also give me some tips and tricks to handle the limits like a pro. How do you tackle such problems? What are the important things one should take care of while solving any problem on limits which prove to be helpful in solving the problem?

NOTE: I’m a high school student.

amWhy
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dRIFT sPEED
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    The usual approach is L'Hospital. If you are not supposed to use something else, you should choose this approach. – Peter Mar 25 '20 at 10:49
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    @paras Khosla no it doesn’t answer my question. By the way that question is asked by me – dRIFT sPEED Mar 25 '20 at 11:01
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    The limit in the link $\lim_{x\to 0}\left(\frac{e}{(1+x)^{1/x}}\right)^{1/x}=\exp\left(\lim_{x\to 0}\frac{e-(1+x)^{1/x}}{x(1+x)^{1/x}}\right)$, these are approached by the same method as answered in the link. And in fact from the previous result, this limit to be calculated follows, or vice versa. – Paras Khosla Mar 25 '20 at 11:05

2 Answers2

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Considering$$y=\frac{e - (1+x)^{\frac{1}{x}}}{x}$$ first, work with $$A=(1+x)^{\frac{1}{x}}\implies \log(A)=\frac 1x\log(1+x)$$ Now, use Taylor series $$\log(A)=\frac1x\left(x-\frac{x^2}{2}+O\left(x^3\right) \right)=1-\frac{x}{2}+O\left(x^2\right)$$ Continue with Taylor $$A=e^{\log(A}=e-\frac{e x}{2}+O\left(x^2\right)$$ Just finish.

Claude Leibovici
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First, note that $$ \lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e^{\lim_{x\rightarrow 0}\frac{1}{x}\ln(1+x)}. $$ After one application of l'Hopital's rule, the exponent becomes $\lim_{x\rightarrow 0}\frac{1}{1+x}$, which equals $1$. This computation can be used to verify that your original limit is, indeed, of the indeterminate form "$\frac{0}{0}$".

Therefore, we use l'Hopital's rule on the original limit to get that the original limit is $$ \lim_{x\rightarrow 0}-\frac{d}{dx}(1+x)^{\frac{1}{x}}. $$ Observe now that $(1+x)^{\frac{1}{x}}=e^{\left(\frac{1}{x}\ln(1+x)\right)}$. Therefore, the derivative is \begin{align*} \frac{d}{dx}(1+x)^{\frac{1}{x}}&=\frac{d}{dx}e^{\left(\frac{1}{x}\ln(1+x)\right)}\\ &=e^{\left(\frac{1}{x}\ln(1+x)\right)}\frac{d}{dx}\left(\frac{1}{x}\ln(1+x)\right)\\ &=e^{\left(\frac{1}{x}\ln(1+x)\right)}\left(-\frac{1}{x^2}\ln(1+x)+\frac{1}{x(x+1)}\right) \end{align*} Therefore, we are interested in calculating $$ \lim_{x\rightarrow 0}e^{\left(\frac{1}{x}\ln(1+x)\right)}\left(\frac{1}{x^2}\ln(1+x)-\frac{1}{x(x+1)}\right) $$

By the calculation above, the exponential part in this limit will approach $e$, so we are left with evaluating the second part, i.e., $$ \lim_{x\rightarrow 0}\left(\frac{1}{x^2}\ln(1+x)-\frac{1}{x(x+1)}\right)=\lim_{x\rightarrow 0}\frac{(1+x)\ln(1+x)-x}{x^2(1+x)}. $$ A substitution of $x=0$ shows that this is also of the indeterminate form "$\frac{0}{0}$", so we apply l'Hopital's rule to get $$ \lim_{x\rightarrow 0}\frac{\ln(1+x)}{x(3x+2)}. $$ Applying l'Hopital's rule one more time to this indeterminate form "$\frac{0}{0}$" gives $$ \lim_{x\rightarrow 0}\frac{1}{(1+x)(6x+2)}=\frac{1}{2}. $$ Now, multiply and complete the calculation.

To learn how to "handle limits like a pro", there is no substitute for just lots of practice. After doing enough, you'll know some tricks and test them out one at a time to see if any of them look promising (again, promising comes from experience).

Michael Burr
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