Limit

**kjchauhan** · October 5th 2010, 09:57 AM

Please help me to prove this:

$\lim_{x\to\alpha} \dfrac{\alpha \sin x - x \sin \alpha} {\alpha \cos x - x \cos \alpha} = \tan (\alpha - \tan^{-1} \alpha)$

Hint is :

$\sin \alpha = \alpha - \frac {\alpha^3}{3!} + ...$
and
$\cos \alpha = 1 - \frac{\alpha^2}{2!} + ...$

Thanks in advance...

**Amer** · October 5th 2010, 11:15 AM

I did not used the hint

$\displaystyle \lim _{x \rightarrow a} \frac{a \sin x - x \cos a}{a\cos x - x \cos a}$

use lopital rule

$\displaystyle \lim _{x\rightarrow a} \frac{a \cos x - \cos a}{-a\sin x - \cos a}= \frac{ a \cos a - \sin a}{-a\sin a - \cos a}$

i want to show that $\displaystyle \tan ( a - \tan ^{-1} a)$

$\tan ( a - \tan ^{-1} a ) = \frac{\sin (a - \tan ^{-1} a)}{\cos (a - \tan^{-1} a)}= \frac{ a \cos a - \sin a}{-a\sin a - \cos a}$

remark that
$tan ^{-1} a = \sin ^{-1} \left( \frac{a}{\sqrt{a^2+1}}\right) = \cos ^{-1} \left( \frac{1}{\sqrt{a^2+1}} \right)$

now
$\displaystyle \frac{\sin (a - \tan ^{-1} a)}{\cos (a - \tan^{-1} a)} = \frac{\sin a\cdot \cos (\tan ^{-1} a) - \cos a \cdot \sin (\tan ^{-1} a) }{\cos a \cdot \cos (\tan ^{-1} a ) + \sin a \sin ( \tan ^{-1} a) }$

simplify u will get $\frac{a \cos a - \sin a}{-a\sin a - \cos a}$

or from beginning

$\displaystyle \tan(a - \tan ^{-1} a ) = \frac{\tan a - a }{1 +a \tan a}$ multiplying with cos a

$\displaystyle \frac{\sin a - a \cos a}{\cos a + a\sin a }$ and this is the limit result after using lopital

**kjchauhan** · October 5th 2010, 07:24 PM

thanks.

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