Find $\lim_{x\to 0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$

Question

If $$f(x)= \frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$$ is continuous at $x=0$ then find $f(0)$

$$ f(0)=\lim_{x\to0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}\\ =\lim_{x\to0}\big[\frac{a^x-1}{x}\big]^3.\frac{x\log a}{\sin(x\log a)}.\frac{x^2}{\log a.\log(1+x^2\log a^2)}\\ =(\log a)^3.1.\lim_{x\to0}\frac{x^2}{\log a.\log(1+x^2\log a^2)}\\ =(\log a)^2.1.\lim_{x\to0}\frac{x^2}{\log(1+x^2\log a^2)}=? $$

Can I use the fact that $\log(1+x)=x-x2/2+x^3/3-....$ but is it not for $-1<x\leq1$ ?

Answer 1

We can use that

$$\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log (a^2))}=\frac{(a^x-1)^3}{x^3}\frac{x^3}{\sin(x\log a)\log(1+x^2\log (a^2))}$$

and by standard limits

$$\frac{(a^x-1)^3}{x^3} \to \log^3 a$$

$$\frac{x^3}{\sin(x\log a)\log(1+x^2\log (a^2))}=\frac{x\log a}{\sin(x\log a)}\frac{x^2\log (a^2)}{\log(1+x^2\log (a^2))}\frac1{2\log^2 a} \to \frac1{2\log^2 a}$$

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Edit

Yes we can also use the series expansion since $x^2\log^2 a \to 0$ and therefore eventually $0<x^2\log (a^2)<1$.

Answer 2

Hint: Use the fact that $\lim_{x\to 0}\frac{\ln(x+1)}{x}=1$.

Answer 3

Using series expansions and $o$-notation you get for $x\to 0$:

$$\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)} = \frac{(e^{x\log a}-1)^3}{(x\log a +o(x))(2x^2\log a+ o(x^3))}$$ $$= \frac{(x\log a+o(x))^3}{2x^3\log^2 a+ o(x^3)} = \frac{x^3\log^3 a+o(x^3)}{2x^3\log^2 a+ o(x^3)}$$ $$=\frac{\log^3 a+o(1)}{2\log^2 a+ o(1)} \stackrel{x\to 0}{\longrightarrow}\frac{\log a}{2}$$