# a limit problem?

$$\lim_{x\to 0}{e^x+ \ln{1-x\over e }\over \tan x-x}$$

i tried doing l'hospital. but couldn't do ! could anyone help ?

• What is the base of the log function? – zoli Jun 13 '15 at 17:42
• the base of the log function is e – saudade Jun 13 '15 at 17:46
• As is often the case, expanding numerator and denominator in a series does the job. We only need up to the $x^3$ terms. – André Nicolas Jun 13 '15 at 17:56
• If you are willing to use L'Hospital's Rule, it is routine calculation, just do it $3$ times. – André Nicolas Jun 13 '15 at 18:01
• Anybody with a answer without lhopital or taylor? Still trying , though my lights went out. :/ Dammit, these powercuts. – Mann Jun 13 '15 at 18:13

This solution uses the standard limits \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}} &=&\frac{1}{6} \\ \lim_{x\rightarrow 0}\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}} &=&-\frac{1}{% 3} \\ \lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3}. \end{eqnarray*} Re-write the original expression as follows \begin{eqnarray*} \frac{e^{x}+\ln \left( \frac{1-x}{e}\right) }{\tan x-x} &=&\frac{e^{x}+\ln (1-x)-\ln e}{\tan x-x} \\ &=&\frac{\left( e^{x} {\color{red}{-1-x-\frac{1}{2}x^{2}}}\right) + {\color{red}{ \left( 1+x+\frac{1}{2}x^{2}\right)}}+\left( \ln (1-x)+ \color{blue}{ x+\frac{1}{2}x^{2}}\right) -\color{blue}{ (x+\frac{1}{2}x^{2}) } -1}{\tan x-x} \\ &=&\frac{\left( e^{x}-1-x-\frac{1}{2}x^{2}\right) +\left( \ln (1-x)+x+\frac{1% }{2}x^{2}\right) }{\tan x-x} \\ &=&\left( \frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}}+\frac{\ln (1-x)+x+\frac{1% }{2}x^{2}}{x^{3}}\right) \left( \frac{x^{3}}{\tan x-x}\right) \end{eqnarray*} passing to the limit one gets \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{e^{x}+\ln \left( \frac{1-x}{e}\right) }{\left( \tan x-x\right) } &=&\lim_{x\rightarrow 0}\left( \frac{e^{x}-1-x-\frac{1}{2}% x^{2}}{x^{3}}+\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}}\right) \left( \frac{% x^{3}}{\tan x-x}\right) \\ &=&\left( \lim_{x\rightarrow 0}\frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}}% +\lim_{x\rightarrow 0}\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}}\right) \left( \lim_{x\rightarrow 0}\frac{x^{3}}{\tan x-x}\right) \\ &=&\left( \frac{1}{6}-\frac{1}{3}\right) \left( \frac{3}{1}\right) \\ &=&-\frac{1}{2}. \end{eqnarray*}

• I loved this answer, and you whoever you are. Just simply beautiful. +10000 – Mann Jun 13 '15 at 19:22
• @Mann. thanks... – Idris Addou Jun 13 '15 at 19:30
• I'm floored. That's just impressive. – FundThmCalculus Jun 13 '15 at 20:33
• Each of those limits you have mentioned can't be derived without the use of any high level techniques like L'Hospital and Taylor. Unfortunately the given problem does not seem to be solvable without the use of these methods. +1 from my end. – Paramanand Singh Jun 14 '15 at 4:03

This screams "Taylor" to me. First, $\log\frac{1-x}e=\log (1-x)-\log e=\log(1-x)-1$. We have $$e^x=1+x+\frac{x^2}2+\frac{x^3}6+o(x^4),\ \ \log(1-x)=-x+\frac{x^2}2-\frac{x^3}3+o(x^4),$$ $$\tan x=x-\frac{x^3}3+o(x^5).$$ So \begin{align} {e^x+ \ln{1-x\over e }\over \tan x-x}&=\frac{1+x+\frac{x^2}2+\frac{x^3}6+o(x^4)+(-x+\frac{x^2}2-\frac{x^3}3+o(x^4))-1}{x-\frac{x^3}3+o(x^5)-x}\\ &=\frac{\frac{x^3}6+o(x^4)}{-\frac{x^3}3+o(x^5)} =\frac{\frac12+o(x)}{-1+o(x)}, \end{align} so the limit as $x\to0$ is $-1/2$.

• Excellent approach! – FundThmCalculus Jun 13 '15 at 17:58
• I benefited from the fact that when I took calculus a million years ago the teacher was a numerical analyst, so he abhorred L'Hôpital. And it pays off, because you approximate your functions with knowledge of how the approximation is working. – Martin Argerami Jun 13 '15 at 18:00
• That's really good. I am an engineer, and I do numerical algorithms, so that method always made a ton of sense to me. I should start using it more often. – FundThmCalculus Jun 13 '15 at 18:01
• Nice to know that all my crazy algebra did work out to be the same answer as yours. :) – FundThmCalculus Jun 13 '15 at 18:12
• One does not need to be a numerical analyst to think that the elevation, in some curricula, of L'Hôpital to a status of THE main method to attack these problems, is, yes, a nuisance. – Did Jun 14 '15 at 11:57

You can change it up a bit, by splitting out the denominator to check for form for L'Hospital's Rule: $$\lim_{x\rightarrow 0} \frac{e^x+\ln \left(\frac{1-x}{e}\right)}{\tan x - x}= \lim_{x\rightarrow 0} \left(e^x+\ln \left(\frac{1-x}{e}\right)\right)*\lim_{x\rightarrow 0} \frac{1}{\tan x - x}$$ The numerator limit is easy, since it exists: $$\lim_{x\rightarrow 0} \left(e^x+\ln \left(\frac{1-x}{e}\right)\right)= \left(e^0 +\ln \left(\frac{1-0}{e} \right)\right)=(1+\ln(e^{-1}))=0$$ The denominator limit is easy, since it exists: $$\lim_{x\rightarrow 0} {\tan x - x}={\tan 0 - 0}={0}$$ Since you have a valid form, you can apply L'Hospitals Rule. For reference, the rule is: $$\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}$$ If and only if the new limit exists!

So you have: $$\lim_{x\rightarrow 0} \frac{e^x+\ln \left(\frac{1-x}{e}\right)}{\tan x - x} = \lim_{x\rightarrow 0} \frac{e^x+\frac{1}{\frac{1-x}{e}}*\frac{-1}{e}}{\sec^2 x -1}$$ $$=\lim_{x\rightarrow 0} \frac{e^x+\frac{-1}{1-x}}{\sec^2 x -1}$$ This is still $\frac{0}{0}$, so repeat again: $$=\lim_{x\rightarrow 0} \frac{e^x+(1-x)^{-2}*-1}{-2*\cos^{-3} x *-\sin x}$$ Apply again: $$=\lim_{x\rightarrow 0} \frac{e^x+-2*(1-x)^{-3}}{6*\cos^{-4} x *\sin^2 x+-2\cos^{-3}x*-\cos x} = \lim_{x\rightarrow 0} \frac{e^x+-2*(1-x)^{-3}}{6*\cos^{-4} x *\sin^2 x+2\cos^{-2}x}$$ This limit exists, and evaluates to: $$\frac{e^0-2*(1-0)^{-3}}{6\cos^{-4} 0 \sin^2 0 +2\cos^{-2} 0}=\frac{1-2}{6/1^4*0^2+2/1^2}=\frac{-1}{2}$$

• I'm working on the solution, my computer quit formatting, so I posted to check the formatting. – FundThmCalculus Jun 13 '15 at 18:04
• Thanks. It wasn't that bad. But nice to know that it is appreciated. – FundThmCalculus Jun 13 '15 at 20:34

Write $\log{1-x\over e}=\log{1-x}-1$ and use Taylor expansions to order $3$

$$e^x+\log{1-x\over e}=1+x+{x^2\over 2}+{x^3\over 6}-x-{x^2\over 2}-{x^3\over 3}-1+o(x^3)$$

And

$$\tan x-x={x^3\over 3}+o(x^3)$$

So

$${e^x+ \ln{1-x\over e }\over \tan x-x}={{-x^3\over 6}+o(x^3)\over {x^3\over 3}+o(x^3)}$$

So the limit we're looking for is $-{1\over 2}$