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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?

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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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$$ \lim_{x\to 2} \frac{e^x - e^2}{x-2} = \left. \frac d {dx} e^x\, \right|_{x=2} = e^2. $$ You could say that's "without L'Hopital's rule" since it's based on the definition of "derivative" plus the subsequent derivation of the derivative of the natural exponential function. Similarly $$ \lim_{x\to2}\frac{(\sin(\pi x/2))^3 - (\sin(\pi\cdot2/2))^3}{x-2} = 0 $$ because that is the value of a derivative. So \begin{align} & \lim_{x\to2} \frac{(\sin(\pi x/2))^3}{e^x-e^2} = \lim_{x\to2} \frac{\left( \frac{(\sin(\pi x/2))^3 - (\sin(\pi\cdot2/2))^3}{x-2} \right)}{\left( \frac{e^x-e^2}{x-2} \right)} \\[10pt] = {} & \text{a quotient of values of two derivatives at }x=2. \end{align}

$x\mapsto \log\left(1 + \frac 1 x\right)$ is no problem: It's continuous at $x=2$, so just plug that in.

I suspect you can deal with the other logarithm in a way similar to what is done above.

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