Evaluate Limit

**Robb** · June 12th 2009, 04:10 AM

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 $\frac{1}{4}$ ;

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

**Amer** · June 12th 2009, 04:16 AM

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

**Robb** · June 12th 2009, 04:23 AM

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?

**Amer** · June 12th 2009, 04:30 AM

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

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

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

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

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

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

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

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

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

**mr fantastic** · June 12th 2009, 04:30 AM

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:

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

Spoiler:

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

**Robb** · June 12th 2009, 04:39 AM

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

**mr fantastic** · June 12th 2009, 04:47 AM

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.

**Amer** · June 12th 2009, 04:56 AM

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

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

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

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

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

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

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

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

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

[tex]

I correct them or not ??

**mr fantastic** · June 12th 2009, 05:11 AM

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

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

I correct them or not ??

Se the red numerator of the last line

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