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Fund the value of $$\lim_{x\rightarrow 0} \frac{e-(1+x)^{\frac{1}{x}}}{\tan x}$$

I have got the answer but by using series expansion and answer is $\frac{e}{2}$, can this be solved by using L.Hospital rule. This question was forwarded to me by a student who dont want to use expansion.

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  • $\begingroup$ Do yo mean that L'Hospital must be used ? Did he try it ? $\endgroup$ – Yves Daoust Oct 7 '17 at 18:24
  • $\begingroup$ Yes and i also tried it and it was getting more and more complicated. At every stage L hospital rule was applied $\endgroup$ – Samar Imam Zaidi Oct 7 '17 at 18:27
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Note that $$\frac{e-(1+x)^{\frac{1}{x}}}{\tan x}= e\cdot \frac{1-\exp\left(\frac{\ln(1+x)}{x}-1\right)}{\left(\frac{\ln(1+x)}{x}-1\right)}\cdot \frac{\ln(1+x)-x}{x^2}\cdot \frac{x}{\tan(x)}.$$ Now recall that $$\lim_{x\to 0}\frac{\tan(x)}{x}=1,\quad \lim_{x\to 0}\frac{\ln(1+x)}{x}=1,\quad \lim_{t\to 0}\frac{e^t-1}{t}=1,$$ and use L'Hospital rule for the remaining factor, $$\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}= \lim_{x\to 0}\frac{1/(1+x)-1}{2x}=\lim_{x\to 0}\frac{-x}{2x(1+x)}=-\frac{1}{2}.$$ Hence, putting all together, we obtain $$\lim_{x\to 0}\frac{e-(1+x)^{\frac{1}{x}}}{\tan x}= e\cdot (-1)\cdot\left(-\frac{1}{2}\right)\cdot 1=\frac{e}{2}.$$

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First of all, $\lim_{x \to 0} \dfrac{\tan x}{x} =1 $, so the denominator can be replaced by $x$.

Then, as $x \to 0$, $\log(1+x) = x-x^2/2+O(x^3)$ and $e^x = 1+x+O(x^2)$ so that

$\begin{array}\\ (1+x)^{1/x} &=e^{\log(1+x)/x}\\ &=e^{(x-x^2/2+O(x^3))/x}\\ &=e^{1-x/2+O(x^2)}\\ &=e\cdot e^{-x/2}e^{O(x^2)}\\ &=e\cdot (1-x/2+O(x^2))(1+O(x^2))\\ &=e\cdot (1-x/2+O(x^2))\\ &=e-ex/2+O(x^2)\\ \text{so that}\\ e-(1+x)^{1/x} &=ex/2+O(x^2)\\ \text{and}\\ \dfrac{e-(1+x)^{1/x}}{x} &=e/2+O(x)\\ \end{array} $

Note: Wolfy agrees, saying the quotient is

$e/2 - (11 e x)/24 + (7 e x^2)/16 - (2447 e x^3)/5760 + (959 e x^4)/2304 + O(x^5) $.

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    $\begingroup$ "who dont want to use expansion." $\endgroup$ – Yves Daoust Oct 7 '17 at 18:35
  • $\begingroup$ You are right. I jumped in too soon. $\endgroup$ – marty cohen Oct 7 '17 at 21:54
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The denominator can be replaced by $x$ (thanks to the limit of $\tan x/x$).

Then the derivative of the numerator is

$$(e-(1+x)^{1/x})'=-(1+x)^{1/x}\left(\frac{\log(1+x)}x\right)'$$ and the first factor tends to $e$.

Then

$$\frac{x-(1+x)\log(1+x)}{x^2(1+x)}$$ is also an indeterminate form, which we can evaluate by two applications of L'Hospital as

$$\frac{1-\log(1+x)-1}{2x+3x^2}\to-\frac{\dfrac1{1+x}}{2+6x}\to-\frac12.$$

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