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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

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  • $\begingroup$ @REr i don't really understand what do you mean $\endgroup$ – user315918 Feb 27 '16 at 23:32
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    $\begingroup$ The limit from the left goes toward $-\infty$ while the limit from the right goes toward $\infty$. $\endgroup$ – Carser Feb 27 '16 at 23:34
  • $\begingroup$ The limit is $\pm\infty$ since the numerator is non-zero at the limit point... $\endgroup$ – abiessu Feb 27 '16 at 23:34
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    $\begingroup$ You can't use l'Hôpital, because the numerator hasn't limit $0$, whereas the denominator has. $\endgroup$ – egreg Feb 28 '16 at 0:11
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$$\lim_{x \to 1^{+}}\frac{x\cdot \sin^2{x}}{x-1}$$

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

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


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

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

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

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

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  • $\begingroup$ $x \to 1$ not just $x \to 1^{+}$ $\endgroup$ – user315918 Feb 27 '16 at 23:38
  • $\begingroup$ The OP changed it, I will edit to put more $\endgroup$ – 3SAT Feb 27 '16 at 23:38
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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 $X\to 0$.

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