Solving limits with powers of x

Question

$$L=\lim_{x \to 0} \frac {3^x + 4^x - 2^{x+1} }{x} $$

Can anyone explain how to solve this and similar questions?

Answer 1

You just need to know the basic limit $$ \lim_{x\to0}\frac{a^x-1}{x}=\log a $$ (which you may prefer to write $\ln a$). Indeed, you have $$ \lim_{x\to0}\frac {3^x + 4^x - 2^{x+1} }{x}= \lim_{x\to0}\left( \frac{3^x-1}{x}+ \frac{4^x-1}{x}- 2\frac{2^x-1}{x} \right) $$ Just rewrite the numerator as $$ 3^x+4^x-2^{x+1}= (3^x-1)+(4^x-1)-(2\cdot2^{x}-2)= (3^x-1)+(4^x-1)-2(2^{x}-1) $$

Answer 2

If you know and are allowed to use Taylor series:

Hint: Start by rewriting $a^x = e^{x\ln a}$. Then, expand each term in the denominator to order 1 (i.e., $e^{x\ln a} = 1+x\ln a + o(x)$ when $x\to 0$) and look for cancellations/factorizations.

Details (place your mouse over the gray area to reveal it):

$$\begin{align} 3^x+4^x - 2^{x+1} &= e^{x\ln 3}+e^{x\ln 4}+ 2\cdot 2^x = e^{x\ln 3}+e^{2x\ln 2}+ 2e^{x\ln 2} \\ &= 1+x\ln 3+1+2x\ln 2-2(1+x\ln 2) + o(x) \\ &= x\ln 3 + o(x) \end{align}$$

so that

$$\begin{align} \frac{3^x+4^x - 2^{x+1}}{x} &= \ln 3 + o(1) \xrightarrow[x\to 0]{} \ln 3 \end{align}$$