Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$

What I attempted thus far:

Multiplying by conjugate

$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$

factor out $\sin x$ in the numerator

$$\lim_{x \to 0} \frac{\sin x \cdot (\sec x - 1)}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$

simplify using $\lim_{x \to 0} \frac{\sin x}{x} = 1 $

$$\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$

From here I don't see any useful direction to go in, if I even went in an useful direction in the first place, I don't know.

I suspect that this could be evaluated using the definition of derivatives, if so, or not, any suggestions?

Solution 1:

Hint: You are doing well. Now multiply top and bottom by $\sec x+1$, and note that $\sec^2 x-1=\tan^2 x$.

Solution 2:

$\begin{align} \lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} &= \lim_{x\to 0}\frac{1}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}\cdot\lim_{x\to 0}\frac{\sec x-1}{x^2}\\ &=\frac12 \cdot \lim_{x\to 0}\frac{\sec x-1}{x^2} \end{align}$

This last limit should yield to L'Hopital's rule, or you could multiply by the conjugate of the numerator and apply a trig identity, plus the fact that $\frac{\tan x}{x}\to 0$ as $x\to 0$.

Solution 3:

If you are allowed to use Taylor series, you could start with $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ Now, use $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Replacing $y$ by the previous developments, you then have $$\sqrt{1+\tan(x)}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{11 x^3}{48}-\frac{47 x^4}{384}+O\left(x^5\right)$$ $$\sqrt{1+\sin(x)}=1+\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{48}+\frac{x^4}{384}+O\left(x^5\right)$$ $$\sqrt{1+\tan(x)}-\sqrt{1+\sin(x)}=\frac{x^3}{4}-\frac{x^4}{8}+O\left(x^5\right)$$ and finally $$\frac{\sqrt{1+\tan(x)}-\sqrt{1+\sin(x)}}{x^3}=\frac{1}{4}-\frac{x}{8}+O\left(x^2\right)$$ which shows the limit and how it is approached when $x$ goes to $0$.

Solution 4:

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

Intermediate calculations: $$ \lim_{x\to0} \frac {2(1 - cos(x))} {x^2} = 1 $$ $$ \lim_{x\to0} cos(x) = 1 $$ $$ \lim_{x\to0} \frac {sin(x)}{x} = 1 $$