$$\lim_{x \to 0^+}{\frac{\sqrt x +\tan^3x+\sqrt x\sin^2x}{x+x^2\cos x-\tan^2x}}.$$
I tried divide both terms by $x^2$ but I didn't get anywhere doing this. Can someone give me some hints on how to solve this limit?
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$\sin x \sim x, \tan x \sim x [x \to 0^+]$, etc. – xbh Nov 12 '18 at 18:32
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HINT
We have that
$$\frac{\sqrt x +\tan^3x+\sqrt x\sin^2x}{x+x^2\cos x-\tan^2x}=\frac{\sqrt x}{x}\frac{1 +\frac{\tan^3x}{\sqrt x}+\sin^2x}{1+x\cos x-\frac{\tan^2x}x}$$
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The second part goes to 1, can you see why? So all is determined by $\frac{\sqrt x}{x}=\frac1{\sqrt x}$. – user Nov 12 '18 at 18:44
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we can rewrite $tan^3xx^{\frac{-1}{2}} $ so it goes to 0 and we should be able to do the same for $tan^2xx^-1$ and so it also goes to 0 right? – Levio Nov 12 '18 at 18:46
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We have that $$\frac{\tan^3x}{\sqrt x}=\frac{\tan^3x}{x^3}\frac{x^3}{\sqrt x}\to ?$$ – user Nov 12 '18 at 18:47
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