2

$$\lim_{x\to0^{+}}\left(\frac{\cos^2x}{x}-\frac{e^x}{\sin x}\right)$$

One can easily calculate this limit by using series expansions for the functions appearing inside the round brackets, yielding $-1$.
My question is: can anyone give another proof of this result without the use of either l'Hospital's rule or series expansions?

Daniele Tampieri
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  • If you combine the difference into a single fraction, then using l'Hopital's rule twice will lead to the answer. However, I agree with Peter Foreman that using series expansions is probably the most straightforward way. – Greg Martin Oct 01 '19 at 23:45
  • Yes. Is there a more clever way of calculating it...? –  Oct 01 '19 at 23:47

4 Answers4

4

Your limit can be found using the series expansions and hence asymptotics of each function near $x=0$. $$\begin{align} \lim_{x\to0^+}\left(\frac{\cos^2{(x)}}{x}-\frac{e^x}{\sin{(x)}}\right) &=\lim_{x\to0^+}\left(\frac{\sin{(x)}\cos^2{(x)}-xe^x}{x\sin{(x)}}\right)\\ &=\lim_{x\to0^+}\left(\frac{\sin{(x)}(1+\cos{(2x)})-2xe^x}{2x\sin{(x)}}\right)\\ &=\lim_{x\to0^+}\left(\frac{(x+o(x^2))(1+1+o(x))-2x(1+x+o(x))}{2x(x+o(x))}\right)\\ &=\lim_{x\to0^+}\left(\frac{2x-2x-2x^2+o(x^2)}{2x^2+o(x^2)}\right)\\ &=\lim_{x\to0^+}\left(\frac{-2x^2+o(x^2)}{2x^2+o(x^2)}\right)\\ &=\lim_{x\to0^+}\left(-1+o(1)\right)\\ &=-1\\ \end{align}$$

Edit: As stated by Greg Martin one can instead use the first terms of the Laurent series expansion of each fraction about zero giving $$\begin{align} \lim_{x\to0^+}\left(\frac{\cos^2{(x)}}{x}-\frac{e^x}{\sin{(x)}}\right) &=\lim_{x\to0^+}\left(\frac{1+o(x)}{x}-\frac{1+x+o(x)}{x+o(x)}\right)\\ &=\lim_{x\to0^+}\left(\frac1x-\frac1x+\frac{o(x)}{x}-\frac{x+o(x)}{x+o(x)}\right)\\ &=\lim_{x\to0^+}\left(-1+o(1)\right)\\ &=-1\\ \end{align}$$

Peter Foreman
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  • Yes, this is how i calculated the limit (-1). –  Oct 01 '19 at 23:44
  • I think it's more natural to simply calculate the Laurent series of the two fractions individually, yielding $\frac1x + O(|x|)$ and $\frac1x+1+O(|x|)$ respectively. – Greg Martin Oct 01 '19 at 23:46
  • Nowhere did you say in the original question that you calculated the limit this way. You did not even describe how you calculated it. – Toby Mak Oct 01 '19 at 23:55
  • I guess you will have to take my word on that. Can you calculate it otherwise...? –  Oct 02 '19 at 00:01
2

$$=\left(\lim_{x\to0}\dfrac{1-\sin^2x}x-\dfrac1{\sin x}\right)-\lim\dfrac{e^x-1}{\sin x}$$

The second limit converges to $1$

For first, either use $\sin x\approx x$ for $x\to0$

Or use Are all limits solvable without L'Hôpital Rule or Series Expansion to find $$\lim_{x\to0}\left(\dfrac1x-\dfrac1{\sin x}\right)$$

lab bhattacharjee
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1

Just some trigonometry and a bit of asymptotic calculus:

  • $\dfrac{\cos^2x}x=\dfrac{1+\cos 2x}{2x}=\dfrac{2-\cfrac{4x^2}2+o(x^2)}{2x}=\dfrac1x-x+o(x);$
  • $\dfrac{\mathrm e^x}{\sin x}\sim_0\dfrac{\mathrm e^x}{x}=\dfrac{1+x+\dfrac{x^2}2+o(x^2)}x=\dfrac1x+1+\dfrac x2+o(x) $,

Therefore $$\frac{\cos^2x}x-\dfrac{\mathrm e^x}{\sin x}=\dfrac1x-x+o(x)-\Bigl(\dfrac1x+1+\dfrac x2+o(x)\Bigr)=-1-\frac{3x}2+o(x)$$

Bernard
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  • Oops! It's getting late here… Fortunately, that doesn't change the limit. Thanks for pointing it! – Bernard Oct 02 '19 at 00:17
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$\begin{array}\\ \dfrac{\cos^2x}{x}-\dfrac{e^x}{\sin x} &=\dfrac{1-\sin^2x}{x}-\dfrac{e^x}{x+O(x^3)}\\ &=\dfrac{1-(x+O(x^3))^2}{x}-\dfrac{1+x+O(x^2)}{x+O(x^3)}\\ &=\dfrac{1-x^2+O(x^3)}{x}-\dfrac{1+x+O(x^2)}{x}(1+O(x^2))\\ &=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)(1+O(x^2))}{x}\\ &=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)}{x}\\ &=\dfrac{-x-x^2+O(x^3)}{x}\\ &=-1-x+O(x^2)\\ &\to -1\\ \end{array} $

marty cohen
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  • This is no different from using l'Hopital's rule. – PierreCarre Oct 02 '19 at 08:03
  • Of course it is. Using the first few terms of Taylor series generally work and does not depend on assumptions about numerator and denominator of the fraction. – marty cohen Oct 02 '19 at 16:50
  • I don't agree. You are still computing derivatives and the number of terms that you need to use in Taylor's expansion just tells you how many times you would have to apply l'Hopital's rule. If the limit exists, the assumptions turn out to be the same. – PierreCarre Oct 03 '19 at 08:14