$\lim_{x \to 1}\frac{x\cdot \sin^2{x}}{x-1}$

Question

Calculate: $$\lim_{x \to 1}\frac{x\cdot \sin^2{x}}{x-1}$$

I don't how to use hopital rule But i tried to take $X=x-1$ so when $x \to 1$ we get $X \to 0$ but i can't find any result

Answer 1

$$\lim_{x \to 1^{+}}\frac{x\cdot \sin^2{x}}{x-1}$$

$$=\lim_{x \to 1^{+}}x\sin^2x\lim_{x \to 1^{+}}\frac{1}{x-1}$$

$$=\sin^2(1)\lim_{x \to 1^{+}}\frac{1}{\underbrace{x-1}_{\to +0}}=\boxed{\infty}$$

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$$\lim_{x \to 1^{-}}\frac{x\cdot \sin^2{x}}{x-1}$$

$$=\lim_{x \to 1^{-}}x\sin^2x\lim_{x \to 1^{+}}\frac{1}{x-1}$$

$$=\sin^2(1)\lim_{x \to 1^{-}}\frac{1}{\underbrace{x-1}_{\to -0}}=\boxed{-\infty}$$

$\Longrightarrow\lim_{x \to 1}\frac{x\cdot \sin^2{x}}{x-1}$ doet not exist

Answer 2

You could have continued with the change of variable $$\frac{x\cdot \sin^2{x}}{x-1}=\frac{(X+1) \sin ^2(X+1)}{X}=\sin ^2(X+1)+\frac{\sin ^2(X+1)}{X}$$ Now, look at what happens when $X\to 0$.