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I need to evaluate $$\lim_{x \to 0} (\frac{\cos x } {x e^{x}}- \frac{1}{x})$$

Using neither L'Hôspitale rule, nor Taylor series...

My try: $$\frac{\cos x}{xe^x}-\frac{1}{x}=\frac{\cos x - e^x}{xe^x}=e^{\ln {\frac {\cos x -e^x}{xe^x}}}=e^{\ln ({\cos x -e^x})-\ln {xe^x}}$$

but it seems that it won't solve the problem.

Any suggestions?

Thank You..

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  • $\begingroup$ maybe express $cos(x)$ in terms of euler identity? $\endgroup$ – dato datuashvili Jan 14 '14 at 8:20
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We have

$$\frac{\cos x } {x e^{x}}- \frac{1}{x}=\frac{1}{e^x}\frac{\cos x-e^x}{x}=$$ so let $$f(x)=\cos x-e^x$$ hence

$$\lim_{x\to0}\frac{\cos x-e^x}{x}=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)=-\sin(0)-e^0=-1$$ finaly $$\lim_{x\to0}\frac{\cos x } {x e^{x}}- \frac{1}{x}=\lim_{x\to0}\frac{1}{e^x}\frac{\cos x-e^x}{x}=-1$$

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  • $\begingroup$ Happy Rid Sami . $\endgroup$ – mrs Jan 14 '14 at 10:48
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The first step you do is correct; then you can notice that $e^x$ at the denominator can't give any problem, so you want to compute $$ \lim_{x\to0}\frac{\cos x-e^x}{x} = \lim_{x\to0}\frac{\cos x-1+1-e^x}{x} = \lim_{x\to0}\frac{\cos x-1}{x}+\lim_{x\to0}\frac{1-e^x}{x} $$ provided both these limits exist (they do).

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When you reduced to the same denominator, you have an error since "$x$" disappears from denominator.

Now, if you use Taylor around $x=0$, your expression write $-1 + x^2/3 - x^3/6$ and you want the limit at $x=0$.

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If you use Taylor, you have $\cos x -1 \approx -\frac{x^2}{2} +o(x^3)$ and $e^{x}-1 \approx x +o(x^2)$, thus $$\frac{\cos x -e^{x}}{xe^{x}} = \frac{\cos x -1}{x^2}\frac{x}{e^{x}} - \frac{e^{x} -1}{x}\frac{1}{e^{x}} \approx -\frac{1}{2}\frac{x}{e^{x}} - \frac{1}{e^{x}} \overset{x\rightarrow 0}{\longrightarrow} -1$$

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See here,

You can write the function after doing the LCM and you will have a function which has a (0/0) form... So, you can easily apply the L'Hospital rule.... and apply the L'Hospital rule for once and see the magic... :-)

This wil be: ( Cos x - e^x )/( xe^x ) ---------------(a),and after applying the L'Hospital rule we will have

 ( - sin x - e^x ) / ( xe^x + e^x ) ------- just apply the differentiation on the numerator and denominator of the fraction for the exprression------(a)

And you can see now we don't have the (0/0) form and we can easily compute the limit by putting the limiting value of x tends to 0. And you will have the answer as ( - 1).

Hope the answer will help you a bit.best of luck.

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