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I make a mistake somewhere but I cannot find where. The answer is supposed to be $e^2$. I think it can be solved with l'Hopital rule, but that is tedious and error-prone. I was looking for a faster way.

$$\begin{align} \lim_{x \to 2} \frac{e^x - e^2 - (x^2 - 4x + 4)\sin (x - 2)}{\sin[(x - 2)^2] + \sin (x-2)} &= \lim_{x \to 2} \frac{e^x - e^2 - (x - 2)^2\sin (x - 2)}{\sin[(x - 2)^2] + \sin (x-2)} =\\ &=\left [ t = x - 2, x \to 2 \implies t \to 0 \right ] =\\ &=\lim_{t \to 0} \frac{e^{t + 2} - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\\ &=\lim_{t \to 0} \frac{e^2 - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\\ &=\lim_{t \to 0} \frac{- t^2(t + o(t))}{t^2 + o(t^2) + t + o(t)} = \tag{1} \\ &=\lim_{t \to 0} \frac{-t^3}{t} = 0 \end{align} $$

I think I may be making a mistake in $(1)$ by either simplifying away the two $e$'s or rewriting the expression with the little-o notation.

rubik
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3 Answers3

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Setting $x-2=h,$

$$\lim_{x \to 2} \frac{e^x - e^2 - (x^2 - 4x + 4)\sin (x - 2)}{\sin[(x - 2)^2] + \sin (x-2)} $$ $$=e^2\lim_{h\to0}\frac{(e^h-1)}{\sin h^2+\sin h}-\lim_{h\to0}\frac{h^2\sin h}{\sin h^2+\sin h}$$

$$\text{Now,}\lim_{h\to0}\frac{e^h-1}{\sin h^2+\sin h}=\frac{\lim_{h\to0}\frac{(e^h-1)}h}{\lim_{h\to0}\frac{\sin h^2}{h^2}h+\lim_{h\to0}\frac {\sin h}h}=\frac1{1\cdot0+1}$$

$$\text{and }\lim_{h\to0}\frac{h^2\sin h}{\sin h^2+\sin h}=\frac{\lim_{h\to0}h^2\lim_{h\to0}\frac{\sin h}h}{\lim_{h\to0}\frac{\sin h^2}{h^2}\lim_{h\to0} h+\lim_{h\to0}\frac{\sin h}h}=\frac{0\cdot1}{1\cdot0+1}=0$$

as $\displaystyle h\to0\implies\sin h\to0\implies\sin h\ne0$ and $\displaystyle\lim_{h\to0}\frac{e^h-1}h=\lim_{h\to0}\frac{\sin h}h=1 $

lab bhattacharjee
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  • Oh, thank you for showing me a new way! But can you also add something related to my mistake? I'm still not very sure about what I did wrong. – rubik Jul 30 '14 at 16:59
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Although it's tempting you cannot lose this $t$. $$\begin{align} &=\lim_{t \to 0} \frac{e^{\color{red}{t} + 2} - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\\ &=\lim_{t \to 0} \frac{e^2 - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\end{align}$$

In general, you do not have the freedom of evaluating specific parts of your limit at once. The term with $e^{t+2}$ includes a denominator that goes to $0$. This is the source of your problem.

Brad
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  • It is true, however, that $$\lim_{t \to 0} \frac{- t^2\sin t}{\sin(t^2) + \sin t} = 0$$ – Brad Jul 30 '14 at 16:55
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Or, expand in powers of t

$$ \lim_{t->0} \frac{e^2(e^t-1)-t^2 \sin t}{\sin t^2 + \sin t} \\ = \lim_{t->0} \frac{e^2(t + t^2/2 + \ldots) - (t^3 - t^4/6 + \ldots)}{t^2 - t^6/6 + \ldots + t-t^3/6+\ldots} \\ = e^2 $$

user_of_math
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