How do I calculate $\lim_{x \to 0}\frac{x^2 - \tan^2(x)}{x^4}$? I just need a hint, not the entirely solution. Thank you in advance!
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1HInt: Difference of squares. – Rene Schipperus Nov 22 '17 at 13:36
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See https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Nov 22 '17 at 13:48
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Can you use L^Hopitlal's rule? – Jose M Serra Nov 22 '17 at 14:29
3 Answers
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Since$$x^2-\tan^2(x)=x^2-\left(x+\frac{x^3}3+o(x^4)\right)^2=-\frac23x^4+o(x^5)\ldots$$
Another possibility:$$\lim_{x\to0}\frac{x^2-\tan^2x}{x^4}=\lim_{x\to0}\frac{x-\tan(x)}{x^3}\times\frac{x+\tan(x)}x.$$
José Carlos Santos
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You should have a $o(x^4)$ in the first equality (or a $O(x^5)$). – Paolo Intuito Nov 22 '17 at 13:44
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@G.S. Thanks. I've edited my answer. What do you think about the new one? – José Carlos Santos Nov 22 '17 at 13:48
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Jose, you corrected the wrong bracket! Once you square you can afford a $o(x^5)$. The other method might be simpler if OP does not know Taylor's expansions. – Paolo Intuito Nov 22 '17 at 13:52
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It is a bit confusing for me using the Taylor's expansion. I can solve it using the latter method. Thank you! – Cosmin Pletosu Nov 22 '17 at 13:52
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Use the Taylor's expansion of the tangent at $0$:
$$\tan(x) = x+\frac{x^3}{3} +o(x^3)\,.$$
Paolo Intuito
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If we don't know the Taylor's expansion of the tangent then we can use the following L^Hospital.
$$\lim_{x\rightarrow0}\frac{x^2-\tan^2x}{x^4}=\lim_{x\rightarrow0}\frac{2x-2\tan{x}\cdot\frac{1}{\cos^2x}}{4x^3}=$$ $$=\lim_{x\rightarrow0}\frac{x\cos^3x-\sin{x}}{2x^3}=\lim_{x\rightarrow0}\frac{\cos^3x-3x\cos^2x\sin{x}-\cos{x}}{6x^2}=$$ $$=-\lim_{x\rightarrow0}\frac{3x\sin{x}\cos{x}+\sin^2x}{6x^2}=-\frac{3}{6}-\frac{1}{6}=-\frac{2}{3}.$$
Michael Rozenberg
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