limit using Taylor's theorem

Question

i need to find the limit of $\lim _{x \to 0}f(x)$

$f(x)=\frac{\left (\sinh \left (x \right ) \right )^{n}-x^{n}}{\left (\sin \left (x \right ) \right )^{n}-x^{n}}$

i tried this $f(x)=\frac{(\frac{\sinh (x)}{x} )^{n}-1}{(\frac{(\sin (x)}{x})^{n}-1}$

and with Taylor's theorem $\lim _{x \to 0}f(x)=\frac{(1+\frac{x^{2}}{6}+\epsilon (x^{3}))^{n}-1}{(1-\frac{x^{2}}{6}+\epsilon (x^{3}))^{n}-1}$ which is also another indeterminate form

Answer 1

Using the binomial theorem at this point, you get $$ \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{1+\frac{nx^2}{6}+\epsilon(x^3)-1}{1-\frac{nx^2}{6}+\tilde\epsilon(x^3)-1} =\lim_{x\rightarrow 0}\frac{\frac{nx^2}{6}+\epsilon(x^3)}{-\frac{nx^2}{6}+\tilde\epsilon(x^3)}. $$ (Here $\epsilon$ and $\tilde\epsilon$ represents functions whose terms all include a factor of $x^k$ where $k>3$.) Now, use the standard technique of dividing the numerator and denominator by the lowest power of $x$ in the denominator to get $$ =\lim_{x\rightarrow 0}\frac{\frac{1}{x^2}\left(\frac{nx^2}{6}+\epsilon(x^3)\right)}{\frac{1}{x^2}\left(-\frac{nx^2}{6}+\tilde\epsilon(x^3)\right)} = \lim_{x\rightarrow 0}\frac{\frac{n}{6}+\epsilon_1(x)}{-\frac{n}{6}+\tilde\epsilon_1(x)}. $$ (Here $\epsilon_1(x)=\frac{1}{x^2}\epsilon(x)$ and similarly for $\tilde\epsilon_1$.) As $x$ approaches $0$, $\epsilon_1(x)$ and $\tilde\epsilon_1(x)$ approach $0$, and you are left with $$ \frac{\frac{n}{6}}{-\frac{n}{6}}=-1. $$

Answer 2

Let's use the usual route of standard limits combined with the use of advanced tools like Taylor series/L'Hospital's Rule if needed.

We have \begin{align} L &= \lim_{x \to 0}\frac{\sinh^{n}x - x^{n}}{\sin^{n}x - x^{n}}\notag\\ &= \lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right)^{n} - 1}{\left(\dfrac{\sin x}{x}\right)^{n} - 1}\notag\\ &= \lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right)^{n} - 1}{\left(\dfrac{\sinh x}{x}\right) - 1}\cdot\dfrac{\left(\dfrac{\sinh x}{x}\right) - 1}{\left(\dfrac{\sin x}{x}\right) - 1}\cdot\dfrac{\left(\dfrac{\sin x}{x}\right) - 1}{\left(\dfrac{\sin x}{x}\right)^{n} - 1}\notag\\ &= \lim_{u \to 1}\frac{u^{n} - 1}{u - 1}\cdot\lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right) - 1}{\left(\dfrac{\sin x}{x}\right) - 1}\cdot\lim_{v \to 1}\frac{v - 1}{v^{n} - 1}\notag\\ &= n\cdot\frac{1}{n}\cdot\lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right) - 1}{\left(\dfrac{\sin x}{x}\right) - 1}\notag\\ &= \lim_{x \to 0}\frac{\sinh x - x}{\sin x - x}\notag\\ &= \lim_{x \to 0}\dfrac{\dfrac{x^{3}}{6} + o(x^{3})}{-\dfrac{x^{3}}{6} + o(x^{3})}\notag\\ &= -1\notag \end{align}