Find $\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$

L'Hospital's rule is not the α and ω of limits computation. First remove the square roots in the numerator: $$\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=\frac{\tan x-x}{(\sqrt{1+\tan x}+\sqrt{1+x})\sin^2x}$$ Now use equivalents:

• $\tan x-x=x+\dfrac{x^3}3+o(x^3)-x$, hence $\;\tan x-x\sim_0 \dfrac{x^3}3$
• $\sqrt{1+\tan x}+\sqrt{1+x}\xrightarrow[x\to 0]{}2$
• $\sin x\sim_0 x$

$$\text{So we have:}\hspace{8em}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}\sim_0\frac{\dfrac{x^3}3}{2x^2}=\frac x6\to 0.\hspace{8em}$$

Using Taylor series we have

$$\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=\frac{\tan x-x}{\sin^2x(\sqrt{1+\tan x}+\sqrt{1+x})}\sim_0\frac{\frac13x^3}{2x^2}=\frac1{6}x$$ so the desired limit is $0$.

$$\lim_{x \to 0} \dfrac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=L$$

Using L'hopital:

$$\lim_{x \to 0} \dfrac{\dfrac{\sec^2x}{2\sqrt{1+\tan x}}-\dfrac{1}{2\sqrt{1+x}}}{2\sin x \cos x}=L$$

Reordering the denominator: $$\lim_{x \to 0} \dfrac{\dfrac{\sec^2x}{2\sqrt{1+\tan x}}-\dfrac{1}{2\sqrt{1+x}}}{\sin 2x }=L$$

Using L'hopital Again

$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x }=L$$

You cannot use l'hopital again because $\lim_{x\to 0} \cos 2x = 1 > 0$

so replacing:

$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x } = \dfrac{\frac{0-\frac{1}{2}}{2}+\frac{1}{4}}{2\cdot 1} = 0$$