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How can I find the limit of this example $$ \lim_{x\to 0}\frac{\sin^{-1}x-2x}{\sin^{-1}x+2\sin(\frac{1}{2}\sin^{-1}x)[3-4\sin^2(\frac{1}{2}\sin^{-1}x)] } $$ I tried using L'hospital but it was very long. Is there any easy trick? please give me some hint.

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3 Answers 3

up vote 2 down vote accepted

what I usually do in such cases: Since $\lim\limits_{x \to 0} \frac{\sin^{-1}x}{x}=1 \implies$ for $ x\to0$ , $ \sin^{-1}x \approx x$, hence subsitute $x$ for $\sin^{-1}x$ and similarly other trigonometric functions' approximations. your problem becomes $\lim\limits_{x\to0} \frac{-x}{x+x(3-x^2)}$ (here I have subsituted $\sin x\approx x$ (which is true as $x \to 0$)), and now you can easily apply L'Hopital's Rule, which evaluates it to $-1/4$.

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I don't think your answer is right,just consider $\frac{sinx-x}{x^{\frac{1}{2}}}$,if you substitute $sinx$ with $x$,the answer is 0. –  89085731 Jun 25 '12 at 12:38
    
we can substitute when x tends to 0, if here limit x tends to 0,then according to my way, answer is coming out to be 0, which is correct indeed. –  Aang Jun 25 '12 at 12:42

All the terms involved are compositions of continuous functions. Bearing this in mind, my hint would be to write $y = \sin ^{-1} x$ and to re-write the limit as a limit in $y$. Then have another go.

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The function $\sin^{-1}x$ is a little hard to work with, mostly because it is less familiar than the sine function. So let $x=\sin(2t)$. Since we are interested in $x$ close to $0$, we can assume that $|2t|\lt \pi/2$. We use $2t$ to cut down on fractions later. Note that $\sin^{-1}x=2t$ and $x=\sin 2t$. So we are interested in $$\lim_{t\to 0}\frac{2t-2\sin 2t}{2t+(2\sin t)(3-4\sin^2 t)}.$$ This looks more doable. Now it seems reasonable to use the fact that $\sin 2t=2\sin t\cos t$. Do some cancellation of $2$'s. We want $$\lim_{t\to 0}\left(\frac{t}{t+(\sin t)(3-4\sin^2 t)}-\frac{2\sin t\cos t}{t+(\sin t)(3-4\sin^2 t)}\right).$$ The expressions still look a little complicated. But the reciprocals are nice! I think you will not have trouble dealing with the limits of $$\frac{t+(\sin t)(3-4\sin^2 t)}{t}\quad\text{and}\quad \frac{t+(\sin t)(3-4\sin^2 t)}{2\sin t\cos t}$$ as $t\to 0$.

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