# Evaluate Limit

• Jun 12th 2009, 05:10 AM
Robb
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 $\frac{1}{4}$;

$\lim_{x\to0}\frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3}$
• Jun 12th 2009, 05:16 AM
Amer
Quote:

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 $\frac{1}{4}$;

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

do you want me to give you a hint or solve it
• Jun 12th 2009, 05:23 AM
Robb
Solving is fine... i tried multiplying out by $\frac{\sqrt{1+\tan{x}}+\sqrt{1+\sin{x}}}{\sqrt{1+\ tan{x}}+\sqrt{1+\sin{x}}}$ but couldn't find anything useful from it?
• Jun 12th 2009, 05:30 AM
Amer
$lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}$

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

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

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

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

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

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

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

Edited :- fix a mistake in the last line thanks to "mr fantastic"
• Jun 12th 2009, 05:30 AM
mr fantastic
Quote:

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 $\frac{1}{4}$;

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

First multiply the numerator and denominator by the conjugate surd $\sqrt{1 + \tan x} + \sqrt{1 + \sin x}$ and simplify:

Spoiler:

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

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

Spoiler:

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

The well known limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ shuold then be applied.
• Jun 12th 2009, 05:39 AM
Robb
Thanks allot for your help, I didn't think to expand the $\tan (x)$ out!
• Jun 12th 2009, 05:47 AM
mr fantastic
Quote:

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

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

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

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

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

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

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

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

[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.
• Jun 12th 2009, 05:56 AM
Amer
Quote:

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

Quote:

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

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

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

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

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

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

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

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

[tex]

I correct them or not ??
• Jun 12th 2009, 06:11 AM
mr fantastic
Quote:

Originally Posted by Amer
Quote:

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
Quote:

Originally Posted by Amer
$lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}$
$lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)$
$lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}$
$lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{(\sqrt{tanx+1}+\sqrt{sinx+1})x^3}$
$lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}$
$lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{x^3(cosx)(\sqrt{tanx+1}+\sqrt{sinx+1})}$
$lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{x^3cos(x)(1+cosx)(\sqrt{tanx+1}+\sq rt{sinx+1})}$
$lim_{x\rightarrow 0}\frac{{\color{red}sin^3 x}}{x^3cos(x)(1+cosx)(\sqrt{tanx+1}+\sqrt{sinx+1}) }=\frac{1}{4}$ since $lim_{x\rightarrow 0} \frac{sinx}{x} =1$
[tex]

I correct them or not ??

Se the red numerator of the last line