My professor gave us a number of limit exercises to solve without using l'Hôpital's rule, there is one that I am having problems with. I think I may of grouped it incorrectly in the ones where we were not allowed to use L'Hôpital.
The limit is the following:
$$\lim_{x \rightarrow 2} \frac{(\sin(\pi x/ 2))^3 \log(1 + 1/x)}{\log(\frac{x^2 - 3x +3}{x-1})(e^x - e^2)} $$
Is it possible to solve this without L'Hôpital?
Answer
Hint: $$\lim_{x \to 2} \frac{(\sin(\pi x/ 2))^3 \log \left(1 + \dfrac1x\right)}{\log\left(\frac{x^2 - 3x +3}{x-1}\right)\left(e^x - e^2\right)}=\lim_{x \to 2} { \log\left(1 + \dfrac1x\right)}\left(\dfrac{\log\left(\frac{x^2 - 3x +3}{x-1}\right)-\log\left(\frac{2^2 - 3\cdot2 +3}{2-1}\right)}{x-2}\right)^{-1}\left(\dfrac{\sin(\pi x/ 2)-\sin(\pi\cdot2/2)}{x-2}\right)^2\left(\sin(\pi x/2)\right)\left(\dfrac{e^x-e^2}{x-2}\right)^{-1}.$$ Recognize the definition of $\dfrac{\mathrm d}{\mathrm dx}\log\left(\frac{x^2 - 3x +3}{x-1}\right)\Bigg|_{x=2}$, $\dfrac{\mathrm d}{\mathrm dx}\sin(\pi x/2)\Bigg|_{x=2}$ and of $\dfrac{\mathrm d}{\mathrm dx}e^x\Bigg|_{x=2}$.
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