Evaluate the limit.

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January 13th 2010, 07:50 AM
General

Evaluate the limit.

Hi
Without using L`Hospital`s Rule, Evaluate the following limit:
$\lim_{x \rightarrow 0} \frac{ \sqrt{1+tanx} - \sqrt{1+sinx} }{x^3}$
January 13th 2010, 08:34 AM
Unbeatable0

I posted the solution in parts. See if you can continue by yourself after every step you see.

$ \lim_{x\rightarrow 0} \frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3} = $

$ = \lim_{x\rightarrow 0} \frac{(\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}})(\sqrt{1+\tan{x}}+\sqrt{1+\sin{x} })}{x^3 (\sqrt{1+\tan{x}}+\sqrt{1+\sin{x}})} = $

Spoiler:

$ =\lim_{x\rightarrow 0}\frac{1}{\sqrt{1+\tan{x}}+\sqrt{1+\sin{x}}} \lim_{x\rightarrow 0} \frac{\tan{x}-\sin{x}}{x^3} = $

Spoiler:

$ = \frac{1}{2}\lim_{x\rightarrow 0} \frac{1}{\cos{x}} \lim_{x\rightarrow 0} \frac{\sin{x}(1-\cos{x})}{x^3} = \frac{1}{2} \lim_{x\rightarrow 0} \frac{\sin{x}(2\sin^2{\frac{x}{2}})}{x^3} = $

Spoiler:

$ = 2\lim_{x\rightarrow 0} \frac{\sin^3{\frac{x}{2}}\cos{\frac{x}{2}}}{x^3} = 2\lim_{x\rightarrow 0} \cos{\frac{x}{2}}\lim_{x\rightarrow 0} \frac{\sin^3{\frac{x}{2}}}{x^3} = $

Spoiler:

$ = \frac{1}{4}\lim_{x\rightarrow 0} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} = $

Spoiler:

$=\frac{1}{4}$
January 13th 2010, 08:36 AM
Drexel28

Quote:

Originally Posted by General

Hi
Without using L`Hospital`s Rule, Evaluate the following limit:
$\lim_{x \rightarrow 0} \frac{ \sqrt{1+tanx} - \sqrt{1+sinx} }{x^3}$

Let $L$ denote the above limit. Clearly $2L=\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}$ (multiply top and bottom by conjugate and seprate limits). So then $2L=\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}=\lim_{x\to 0}\frac{x+\tfrac{x^3}{3}+\cdots-\left(x-\tfrac{x^3}{6}+\cdots\right)}{x^3}=\lim_{x\to0}\fr ac{\tfrac{x^3}{2}+\cdots}{x^3}=\frac{1}{2}$ so that $L=\frac{1}{4}$
January 13th 2010, 08:41 AM
General

Quote:

Originally Posted by Drexel28

Let $L$ denote the above limit. Clearly $2L=\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}$ (multiply top and bottom by conjugate and seprate limits). So then $2L=\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}=\lim_{x\to 0}\frac{x+\tfrac{x^3}{3}+\cdots-\left(x-\tfrac{x^3}{6}+\cdots\right)}{x^3}=\lim_{x\to0}\fr ac{\tfrac{x^3}{2}+\cdots}{x^3}=\frac{1}{2}$ so that $L=\frac{1}{4}$

Its 2 or $\sqrt 2$ ?
January 13th 2010, 08:43 AM
Drexel28

Quote:

Originally Posted by General

Its 2 or $\sqrt 2$ ?

I misread the question, and made a typo. It should be two. Although, I would suggest you look at Unbeatable's solution. He didn't take the easy way out and use Maclaurin.
January 13th 2010, 08:47 AM
General

What is the general formula of Mac. series for tanx?
i.e: the formula contains Sigma
I know it for arctan x
not tan x
January 13th 2010, 08:53 AM
Drexel28

Quote:

Originally Posted by General

What is the general formula of Mac. series for tanx?
i.e: the formula contains Sigma
I know it for arctan x
not tan x

It's crazy and involves Bernoulli numbers. It's not worth remembering, but just do the calculation three times (the second one vanishes) and remmeber that.