# Calculate limit $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule

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 $\frac{0}{0}$

• Hints: $b^x = e^{\ln(b) x}$ and $\lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1$ – JoDraX Dec 26 '16 at 19:42

## 3 Answers

$\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$

Putting $x = y+2$,

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

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

Put $x=t+2$ and compute

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

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

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

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

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 $\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}=1$.

• 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 – Mark Viola Dec 27 '16 at 2:12
• 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 :) – Dave Dec 27 '16 at 6:51