1. ## Evaluate Limit

Hello,
Doing some revision, and I cant figure out what to do with this function in order to calculate the limit... I know it equals $\displaystyle \frac{1}{4}$;

$\displaystyle \lim_{x\to0}\frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3}$

2. Originally Posted by Robb
Hello,
Doing some revision, and I cant figure out what to do with this function in order to calculate the limit... I know it equals $\displaystyle \frac{1}{4}$;

$\displaystyle \lim_{x\to0}\frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3}$
do you want me to give you a hint or solve it

3. Solving is fine... i tried multiplying out by $\displaystyle \frac{\sqrt{1+\tan{x}}+\sqrt{1+\sin{x}}}{\sqrt{1+\ tan{x}}+\sqrt{1+\sin{x}}}$ but couldn't find anything useful from it?

4. $\displaystyle lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}$

$\displaystyle lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)$

$\displaystyle lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}$

$\displaystyle lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{2x^3}$

$\displaystyle lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{2x^3}$

$\displaystyle lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{2x^3(cosx)}$

$\displaystyle lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{2x^3(1+cosx)}$

$\displaystyle lim_{x\rightarrow 0}\frac{(sin^3x)}{4x^3}=\frac{1}{4}$ since $\displaystyle lim_{x\rightarrow 0} \frac{sinx}{x} =1$

Edited :- fix a mistake in the last line thanks to "mr fantastic"

5. Originally Posted by Robb
Hello,
Doing some revision, and I cant figure out what to do with this function in order to calculate the limit... I know it equals $\displaystyle \frac{1}{4}$;

$\displaystyle \lim_{x\to0}\frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3}$
First multiply the numerator and denominator by the conjugate surd $\displaystyle \sqrt{1 + \tan x} + \sqrt{1 + \sin x}$ and simplify:

Spoiler:

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

Then multiply the numerator and denominator by $\displaystyle 1 + \cos x$ and simplify:

Spoiler:

$\displaystyle \frac{\sin^3 x}{x^3 \cos x \left( \sqrt{1 + \tan x} + \sqrt{1 + \sin x} \right)}$

The well known limit $\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ shuold then be applied.

6. Thanks allot for your help, I didn't think to expand the $\displaystyle \tan (x)$ out!

7. Originally Posted by Amer
$\displaystyle lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}$

$\displaystyle lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)$

$\displaystyle lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}$

$\displaystyle lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{2x^3}$

$\displaystyle lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{2x^3}$

$\displaystyle lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{2x^3(cosx)}$

$\displaystyle lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{2x^3(1+cosx)}$

$\displaystyle lim_{x\rightarrow 0}\frac{sinx(sin^3x)}{4x^3}=\frac{1}{4}$ since $\displaystyle lim_{x\rightarrow 0} \frac{sinx}{x} =1$

[tex]
There are some minor corrections that need to be made in this solution. Rather than point them out, it will be instructive for the interested reader to make them him/herself.

8. Originally Posted by mr fantastic
There are some minor corrections that need to be made in this solution. Rather than point them out, it will be instructive for the interested reader to make them him/herself.
I think I should not sub zero in the limit before I finish the problem like this

Originally Posted by Amer
$\displaystyle lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}$

$\displaystyle lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)$

$\displaystyle lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}$

$\displaystyle lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{(\sqrt{tanx+1}+\sqrt{sinx+1})x^3}$

$\displaystyle lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}$

$\displaystyle lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{x^3(cosx)(\sqrt{tanx+1}+\sqrt{sinx+1})}$

$\displaystyle lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{x^3cos(x)(1+cosx)(\sqrt{tanx+1}+\sq rt{sinx+1})}$

$\displaystyle lim_{x\rightarrow 0}\frac{sinx(sin^3x)}{x^3cos(x)(1+cosx)(\sqrt{tanx +1}+\sqrt{sinx+1})}=\frac{1}{4}$ since $\displaystyle lim_{x\rightarrow 0} \frac{sinx}{x} =1$

[tex]
I correct them or not ??

9. Originally Posted by Amer
Originally Posted by mr fantastic
There are some minor corrections that need to be made in this solution. Rather than point them out, it will be instructive for the interested reader to make them him/herself.
I think I should not sub zero in the limit before I finish the problem like this
Originally Posted by Amer
$\displaystyle lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}$
$\displaystyle lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)$
$\displaystyle lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}$
$\displaystyle lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{(\sqrt{tanx+1}+\sqrt{sinx+1})x^3}$
$\displaystyle lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}$
$\displaystyle lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{x^3(cosx)(\sqrt{tanx+1}+\sqrt{sinx+1})}$
$\displaystyle lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{x^3cos(x)(1+cosx)(\sqrt{tanx+1}+\sq rt{sinx+1})}$
$\displaystyle lim_{x\rightarrow 0}\frac{{\color{red}sin^3 x}}{x^3cos(x)(1+cosx)(\sqrt{tanx+1}+\sqrt{sinx+1}) }=\frac{1}{4}$ since $\displaystyle lim_{x\rightarrow 0} \frac{sinx}{x} =1$
[tex]
I correct them or not ??
Se the red numerator of the last line