Calculate limit with or without the use of l'Hospital's rule or series expansions

Question

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

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?

Answer 1

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 $$\begin{align} \lim_{x\to0^+}\left(\frac{\cos^2{(x)}}{x}-\frac{e^x}{\sin{(x)}}\right) &=\lim_{x\to0^+}\left(\frac{1+o(x)}{x}-\frac{1+x+o(x)}{x+o(x)}\right)\\ &=\lim_{x\to0^+}\left(\frac1x-\frac1x+\frac{o(x)}{x}-\frac{x+o(x)}{x+o(x)}\right)\\ &=\lim_{x\to0^+}\left(-1+o(1)\right)\\ &=-1\\ \end{align}$$

Answer 2

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

Answer 3

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

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 $$\lim_{x\to0}\left(\dfrac1x-\dfrac1{\sin x}\right)$$

Answer 4

$\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} $