Evaluating the $\lim_{x\to0}\frac{4^x-1}{8^x-1}$ without L'Hopital Rule

Question

How to evaluate $\displaystyle \lim_{x\to0}\left(\frac{4^x-1}{8^x-1}\right)$ without L'Hopital rule?

When I evaluated this limit I got an indetermination, $\frac{0}{0}$. I learned that in a rational function when one get $\frac{0}{0}$ indeterminated form, one should look for the common terms between numerator and denominator by factoring. But I can't figure out how to find the common terms in this case. Can you help me? Thanks.

Answer 1

Note that $$4^x-1=(2^2)^x-1=(2^x)^2-1=\color{red}{(2^x-1)}(2^x+1)$$ $$8^x-1=(2^3)^x-1=(2^x)^3-1=\color{red}{(2^x-1)}(2^{2x}+2^x+1)$$

Answer 2

Hint We can factor $a^{k x} - 1$ as: $$a^{k x} - 1 = (a^x - 1)(a^{(k - 1) x} + a^{(k - 2) x} + \cdots + a^x + 1).$$

Answer 3

HINT:

As $\lim_{h\to0}\dfrac{e^h-1}h=1,$ and as $ a=e^{\ln(a)}$

$$\lim_{h\to0}\dfrac{a^h-1}h=\ln(a)\cdot\lim_{h\to0}\dfrac{e^{h\ln(a)}-1}{h\ln a}=\ln a$$

Now we know $\ln(b^m)=m\ln(b)$ when both logarithm remain defined

For real calculus we need $a>0$