Limit using Taylor series expansion

Question

Can someone help me with this limit. I know I have to expand in Taylor series $\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+o(x^3)$ $a^x=e^{x\ln a}=1+x\ln a+\frac{1}{2!}x^2(\ln a)^2+\frac{1}{3!}x^3(\ln a)^3+o(x^3)$. But how to proceed next?

Answer 1

Note that from Taylor's expansion

$$\frac{ a^{\sqrt{x+1}}-a^{1+\frac{x^2}2-\frac{x^2}8} }{x^3} =\left(a^{1+\frac{x^2}2-\frac{x^2}8}\right)\frac{ a^{\frac{x^3}{16}+o(x^3)} -1 }{x^3} =\left(a^{1+\frac{x^2}2-\frac{x^2}8}\right)\frac{ a^{\frac{x^3}{16}+o(x^3)} -1 }{\frac{x^3}{16}+o(x^3)}\frac{\frac{x^3}{16}+o(x^3)}{x^3}$$

now use standard limits.

Answer 2

$$\lim_{ x \to 0} \frac{a^{\sqrt{1+x}}-a^{1+x/2-x^2/8}}{x^3}=\lim_{ x \to 0} \;\underbrace{a^{1+x/2-x^2/8}}_{\to a} \cdot\left(\frac{a^{\sqrt{1+x}-1-x/2+x^2/8}-1}{x^3}\right)$$

Using Taylor series $\displaystyle \sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+\mathcal{O}(x^3)$ ;

$$\lim_{ x \to 0} \;\underbrace{a^{1+x/2-x^2/8}}_{\to a} \cdot\left(\frac{a^{\sqrt{1+x}-1-x/2+x^2/8}-1}{x^3}\right)=a \lim_{ x \to 0} \left(\frac{a^{x^3/16}-1}{x^3}\right)=\frac{a \ln a}{16}$$