Finding $\lim_{x\rightarrow 0}\frac{x^{6000}-(\sin x)^{6000}}{x^{6002}}$

Question

Finding $\displaystyle \lim_{x\rightarrow 0}\frac{x^{6000}-(\sin x)^{6000}}{x^{6002}}$

Try: I have tried using series expansion of $\displaystyle \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots $

So $$\lim_{x\rightarrow 0}\frac{x^{6000}-\bigg(x-\frac{x^3}{3!}+\cdots \bigg)^{6000}}{x^{6002}}$$

$$\lim_{x\rightarrow 0}\frac{x^{6000}-x^{6000}\bigg[1-\frac{x^2}{6}+\cdots \bigg]^{6000}}{x^{6002}} = 1000$$

Could some help me how to solve it without Series expansion or D L Hopital Rule

Thanks

Answer 1

For integer $n\ge0,$

$$\dfrac{x^{n+1}-\sin^{n+1}x}{x^{n+3}}=\dfrac{x-\sin x}{x^3}\cdot\sum_{r=0}^n\dfrac{x^r\sin^{n-r}x}{x^n}$$

Now $$\dfrac{x^r\sin^{n-r}x}{x^n}=\left(\dfrac{\sin x}x\right)^{n-r}$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion

Answer 2

Hint: $\;\dfrac{x^{6000}-(\sin x)^{6000}}{x^{6002}} = \dfrac{1-\left(\dfrac{\sin x}{x}\right)^{6000}}{x^2}=\dfrac{1 - \dfrac{\sin x}{x}}{x^2}\cdot \left(1 + \dfrac{\sin x}{x} + \ldots\right)\,$