Question

(sqrt(1 + tan x) - sqrt(1 +sin x))/x^3

Answer 1

Please find the derivative as x approaches zero

Answer 2

Do you mean find the limit as \(x\to0\), or find the derivative and evaluate it at 0?

Answer 3

The first one

Answer 4

\[\lim_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}\\
\lim_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}\cdot\frac{\sqrt{1+\tan x}+\sqrt{1+\sin x}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\\
\lim_{x\to0}\frac{(1+\tan x)-(1+\sin x)}{x^3\left(\sqrt{1+\tan x}+\sqrt{1+\sin x}\right)}\\
\lim_{x\to0}\frac{\tan x-\sin x}{x^3\left(\sqrt{1+\tan x}+\sqrt{1+\sin x}\right)}\\
\lim_{x\to0}\frac{\sin x\left(\dfrac{1}{\cos x}-1\right)}{x^3\left(\sqrt{1+\tan x}+\sqrt{1+\sin x}\right)}\\
\lim_{x\to0}\frac{\sin x}{x}\cdot\lim_{x\to0}\frac{\dfrac{1}{\cos x}-1}{x^2 \left(\sqrt{1+\tan x}+\sqrt{1+\sin x}\right)}\]
Here's a preliminary step. Not sure yet if it's in the right direction.

Answer 5

@SithsAndGiggles Thanks, that really helped.