Evaluating $\lim\limits_{x\to 1}\frac{2^{\sin(\pi x)}-2^{\tan(\pi x)}}{x-1}$

Question

I am stuck at evaluating this limit: $$\lim\limits_{x \to 1}\frac{2^{\sin(\pi x)}-2^{\tan(\pi x)}}{x-1}$$ Can someone help me, please?

Also, I am not supposed to use L'Hospital's rule or derivatives. Thanks!

Answer 1

How about this.

\begin{eqnarray} \mathcal L &=&\lim_{x\to 1} \frac{2^{\sin \pi x}-2^{\tan \pi x}}{x-1}=\\ &\stackrel{x-1 = t}{=}&\lim_{t\to 0} \frac{2^{-\sin \pi t}-2^{\tan \pi t}}{t}=\\ &=&\lim_{t\to 0} \left(\frac{2^{-\sin \pi t}-1}{t}-\frac{2^{\tan \pi t}-1}{t}\right)=\\ &=&\lim_{t\to 0} \left(\frac{e^{-\sin\pi t\log 2}-1}{t}-\frac{e^{\tan\pi t\log 2}-1}{t}\right)=\\ &=&\lim_{t\to 0}\left[\frac{e^{-\sin\pi t\log 2}-1}{-\sin\pi t\log 2}\cdot \left(-\frac{\sin\pi t \log 2}t\right)-\frac{e^{\tan\pi t\log 2}-1}{\tan \pi t \log 2}\cdot \left(\frac{\tan \pi t \log 2}{t}\right)\right]=\\ &=&\lim_{t\to 0}\left(-1\cdot \pi \log 2-1\cdot \pi \log 2\right)=\\ &=&-2\pi \log 2, \end{eqnarray} where we simply used fundamental limits, i.e. $$\frac{e^{\alpha(x)}-1}{\alpha(x)} \to 1,$$ and $$\frac{\sin\alpha(x)}{\alpha(x)} \to 1$$ when $\alpha(x) \to 0$.

Answer 2

You can then replace $x = y+1$ which gives you $$\lim\limits_{y \to 0}\frac{2^{-\sin(\pi y)}-2^{\tan(\pi y)}}{y}$$

Now, convert numerator in the power of e, then expand using this formula. You can then easily evaluate.