How do I solve this limit without L'Hospital?

Question

$$\lim_{x\to0}\frac{6^x-1}{x}$$

I have spent quite a lot of time on this limit but I still can't solve it. None of the regular tricks work here, I can't get rid of the $x$, nor can I get it in the form $$\left(1+\left(\frac1y\right)\right)^y.$$ So, how can I solve this limit without using L'Hospital's Rule?

Noting that $6^0=1$, can you turn this into a derivative and solve it? — BHT, Mar 14 '20 at 17:45
@DietrichBurde: I think one can't forbid its usage in a term end exam. Some instructors do however forbid in some routine tests/assignments while one is learning the basics of limits. — Paramanand Singh, Mar 14 '20 at 18:15
The proper answer is based on a proper definition of symbol $6^x$ and in general on that of $a^x$. — Paramanand Singh, Mar 14 '20 at 18:16
A typical approach is that one is first made aware of functions $\exp(x) $ and $\log x$ and the limit $\lim_{x\to 0}\dfrac{\exp(x)-1}{x}=1$. Once this is done one can define $a^x=\exp(x\log a) $ and then one easily obtains $(a^x-1)/x\to \log a$ as $x\to 0$. — Paramanand Singh, Mar 14 '20 at 18:20

score 0 · Answer 1 · answered Mar 14 '20 at 18:31

$$\frac {6^x-1}{x}=\frac {e^{(\ln 6)x}-1}{x}$$ $$=\frac{(1+(\ln 6)x+\frac {1}{2!}(\ln 6)^2x^2+...)-1}{x}$$ $$=\ln 6+\frac {1}{2!}(\ln 6)^2x+...$$ Thus, taking the limit as $x \rightarrow 0$ and using the fact that a power series is continuous, the desired limit is ln6.

How do I solve this limit without L'Hospital?

1 Answers1