difficult trig limit: [(sqrt(1+tan x)-sqrt(1-sinx))/x^3]

[xXAceFireXx]

lim(x->0) [(sqrt(1 + tan x) - sqrt(1 - sinx)) / x^3]

I tried multiplying the top and bottom by the conjugate, giving me tan x - sin x on top, but I don't know where to go from there.

Thank you for your help!

 

[galactus]

Re: difficult trig limit...

\(\displaystyle \L\\\lim_{x\to\0}\frac{\sqrt{1+tan x}-\sqrt{1-sinx}}{x^{3}}\)

i tried multiplying the top and bottom by the conjugate, giving me tan x - sin x on top, but i dont know where to go from there.

thanks for the help

After doing the conjugate thing, the numerator becomes tan(x)+sin(x)

Then you have:

\(\displaystyle \L\\\frac{tan(x)+sin(x)}{x^{3}(\sqrt{1+tan(x)}+\sqrt{1-sin(x)})}\)

Rewrite as:

\(\displaystyle \L\\\left(\frac{sin(x)}{x^{3}}+\frac{tan(x)}{x^{3}}\right)\cdot\frac{1}{\sqrt{1+tan(x)}+\sqrt{1-sin(x)}}\)

You have:

\(\displaystyle \L\\\left(\lim_{x\to\0}\frac{sin(x)}{x^{3}}+\lim_{x\to\0}\frac{tan(x)}{x^{3}}\right)\cdot\underbrace{\lim_{x\to\0}\frac{1}{\sqrt{1+tan(x)}+\sqrt{1-sin(x)}}}_{\text{limit=1/2}}\)

Therefore, you have:

\(\displaystyle \L\\\left(\lim_{x\to\0}\frac{sin(x)}{x^{3}}+\lim_{x\to\0}\frac{tan(x)}{x^{3}}\right)(\frac{1}{2})\)

=\(\displaystyle ({\infty}+{\infty})(\frac{1}{2})={\infty}\)

If you want to step through the limits of \(\displaystyle \frac{sin(x)}{x^{3}} \;\ and \;\ \frac{tan(x)}{x^{3}}\), you can do that. OK?.