Sunday, 8 March 2015

limits - How to evaluate $lim_{xto0}frac{sin^2left(frac x2right)-frac{x^2}4}{e^{x^2}+e^{-x^2}-2}$?

$$\begin{align*}
\lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e^{-x^{2}}} \\
&= \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} -2xe^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{(2x)(2x)e^{x^{2}} - (2x)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{4x^2 e^{x^{2}} + 4x^2 e^{-x^{2}}} \\

&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\left( -\sin \frac{x}{2} \cos \frac{x}{2} \right) - \frac{1}{2} \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)}{(4x^2)(2x)e^{x^{2}} + (4x^2)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{-\sin \frac{x}{2} \cos \frac{x}{2}}{8x^3e^{x^{2}} - 8x^3 e^{-x^{2}}} \\
\end{align*}$$



After evaluating the limit as $x \to 0$, I noticed that the problem comes up to be in an indeterminate form of $0/0$. I immediately utilized the L'Hospital Rule by differentiating both the numerator and denominator.



However, after using L'Hospital rule for 5-6 times, I noticed that the question will go through a loop of $0/0$ indeterminants.



In my second attempt,
I have tried multiplying $\exp(x^2)$ in both the numerator and denominator with hopes to balance out the $\exp(x^{-2})$. However, an indeterminant is $0/0$ still resulting.




Any help would be appreciated, thank you all!

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...