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I have to find the $$\lim_{x \to 0}\frac{(e^{2\tan(x)}-1) \cdot \ln(2-\cos^2(x))}{\sqrt{1+x^3}-(\cos x)}$$ using only notable limits I managed to solve it having as result $4$ using notable limits, it is quite simple resolve but the solution is not $4$... I can't find any other way of solution, can somebody solve it using notable limits clearly? (I'm asking many questions regarding the calculus of limit because I'm preparing for a limit's test)

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Verify that my edit has not distorted your question? –  Zhoe Nov 21 '13 at 17:59
    
cosx is not under square root –  Dipok Nov 21 '13 at 18:16
    
I think this would be $0$ by L'Hôpital's rule. The numerator will be always $0$ because $e^{2\tan{x}} -1$ term and denominator is $1$ at second derivative. –  plhn Nov 21 '13 at 18:18
    
Unfortunately I can't use derivatives... –  Dipok Nov 21 '13 at 18:28
    
That's kinda tricky, no derivatives...I was also going to suggest L'Hopitals.. –  Zhoe Nov 21 '13 at 18:31

1 Answer 1

up vote 2 down vote accepted

$$\lim_{x \to 0}\frac{(e^{2\tan x}-1) \cdot \ln(2-\cos^2x)}{\sqrt{1+x^3}-\cos x}=\lim_{x \to 0}\frac{e^{2\tan x}-1}{2\tan x}\cdot\lim_{x \to 0}\frac{ \ln(1+\sin^2x)}{\sin^2x}\cdot\lim_{x \to 0}\frac{2\tan x\cdot\sin^2x}{\sqrt{1+x^3}-\cos x}=2\cdot1\cdot\lim_{x \to 0}\frac{\tan x}{x}\lim_{x \to 0}\cdot\frac{\sin^2 x}{x^2}\cdot\lim_{x \to 0}\frac{x^3}{\sqrt{1+x^3}-1+(1-\cos x)}=2\lim_{x \to 0}\frac{x^3}{\frac{x^3}{\sqrt{1+x^3}+1}+(1-\cos x)}= 2\lim_{x \to 0}\frac{1}{\frac{1}{\sqrt{1+x^3}+1}+\frac{1-\cos x}{x^3}}=\frac{2}{\frac{1}{2}+\lim_{x \to 0}\frac{1-\cos x}{x^2}\cdot\frac{1}{x}}=\frac{2}{\frac{1}{2}+\frac{1}{2}\cdot\lim_{x \to 0}\frac{1}{x}}.$$ If we see it as the lateral limit, then the limit is equal to $0.$

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I could not make out your last line. As you have very clearly and beautifully demonstrated the limit exists and is $0$. To be more specific the denominator tends to $-\infty$ as $x \to 0^{-}$ and to $\infty$ as $x \to 0^{+}$. But since the numerator is fixed at $2$, the fraction tends to $0$ in both cases when $x \to 0^{-}$ and $x \to 0^{+}$. Hence the limit exists. –  Paramanand Singh Nov 22 '13 at 6:14

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