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How to calculate limit: $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule?

If $x = 2$, I get uncertainty $\frac{0}{0}$

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    $\begingroup$ Hints: $b^x = e^{\ln(b) x}$ and $\lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1$ $\endgroup$ – JoDraX Dec 26 '16 at 19:42
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$\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)} $

Putting $x = y+2$,

$\begin{array}\\ \dfrac{(2^x)-4}{\sin(\pi x)} &=\dfrac{(2^{y+2})-4}{\sin(\pi (y+2))}\\ &=4\dfrac{(2^{y})-1}{\sin(\pi y)} \qquad\text{since } 2^{y+2} = 4\cdot 2^y \text{ and }\sin(\pi (y+2))=\sin(\pi y + 2\pi)=\sin(\pi y)\\ &=4\dfrac{e^{y\ln 2}-1}{\sin(\pi y)}\\ &\approx 4\dfrac{y\ln 2}{\pi y} \qquad\text{since } e^z \approx 1+z \text{ and } \sin(z) \approx z \text{ for small }z\\ &= \dfrac{4\ln 2}{\pi } \end{array} $

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  • $\begingroup$ Why $\sin(\pi (y+2)) / 4 = \sin(\pi y) $? $\endgroup$ – Dave Dec 26 '16 at 19:54
  • $\begingroup$ Because $\sin(z+2\pi) = \sin(z)$ for all $z$. The 4 doesn't come into this part. $\endgroup$ – marty cohen Dec 26 '16 at 19:58
  • $\begingroup$ Trigonometric formulas, sorry, I not right away understand it. $\endgroup$ – Dave Dec 26 '16 at 20:03
  • $\begingroup$ What is the symbol $\approx$. $\endgroup$ – hamam_Abdallah Dec 26 '16 at 20:10
  • $\begingroup$ Approximately equal to. $\endgroup$ – marty cohen Dec 26 '16 at 23:53
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Put $x=t+2$ and compute

$$4\lim_{t\to 0}\frac{e^{t\ln(2)}-1}{\sin(\pi t)}$$

$$=\frac{4\ln(2)}{\pi}\lim_{t\to 0}\frac{e^{t\ln(2)-1}}{t\ln(2)}\frac{\pi t}{\sin(\pi t)}$$

$$=\frac{4\ln(2)}{\pi}.$$

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  • $\begingroup$ @Fatima, you are right, but I do not understand the solution progress :( $\endgroup$ – Dave Dec 26 '16 at 19:56
  • $\begingroup$ @divisor What didn't you uderstdand. $\endgroup$ – hamam_Abdallah Dec 26 '16 at 20:11
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HINT:

$$\frac{2^x-4}{\sin(\pi x)}=\left(\frac{2^x-4}{x-2}\right)\,\left(\frac{x-2}{\sin(\pi (x-2))}\right) \tag 1$$

The limit of the first parenthetical term in $(1)$ is the derivative of $2^x$ at $x=2$. And $\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}=1$.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy Holidays! -Mark $\endgroup$ – Mark Viola Dec 27 '16 at 2:12
  • $\begingroup$ The problem is that your answer is difficult for me. I upvote your answer. I also try to explore your other answers in gratitude for your time, and upvote their too. Thank you and Happy Holidays :) $\endgroup$ – Dave Dec 27 '16 at 6:51

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