How to evaluate this limit without using L'Hospital's rule

Question

$$\lim_{x\to 0}{(\sin(x)-x+{1 \over 6}x^3)({e^x-1})\over x^6}$$

How to determine this limit without L'Hopital's rule? Is there anyone would like to give some hints.

Should I use power series?

Answer 1

Hint

$$e^x=1+x+{x^2\over 2}+{x^3\over 6}+\cdots$$and $$\sin x=x-{x^3\over 6}+{x^5\over 120}-\cdots$$

Answer 2

Since$$\sin x=x-\frac1{3!}x^3+\frac1{5!}x^5-\cdots$$and$$e^x-1=x+\frac1{2!}x^2+\frac1{3!}x^x+\cdots,$$you have$$\left(\sin(x)-x+\frac16x^3\right)(e^x-1)=\left(\frac1{120}x^5+\cdots\right)\left(x+\frac12x^2+\cdots\right)=\frac1{120}x^6+\cdots$$and therefore your limit is $\frac1{120}$.

Answer 3

$$\sin(x) = x - \frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7\cdots$$

So the limit reduces to $$\lim_{x\to0}\frac{(\frac{1}{5!}x^5-\frac{1}{7!}x^7\cdots)(e^x-1)}{x^6} = \lim_{x\to0}\frac{(\frac{1}{5!}-\frac{x^2}{7!}\cdots)(e^x-1)}{x} = \frac{1}{5!} = \frac{1}{120}$$