Limits - Evaluate without L'Hospital's rule or Series expansion?

**fardeen_gen** · June 9th 2009, 07:51 AM

Evaluate without L'Hospital's rule or Series Expansion:

$\lim_{x\rightarrow 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}$

**Ruun** · June 9th 2009, 08:45 AM

$cos^4 (\alpha) - sin^4 (\alpha) =(cos^2(\alpha) + sin^2(\alpha))(cos^2(\alpha)-sin^2(\alpha))=(cos^2(\alpha)-sin^2(\alpha))$

As $cos(2\alpha)=cos^2(\alpha)-sin^2(\alpha)$ the numerator is zero, and the denominator tends to zero... In this case the limit will be zero. But as x is not four, but tends to four, I'm not sure if the above trig. expressions can be used.

**Moo** · June 9th 2009, 09:55 AM

Hello,

Originally Posted by Ruun

$cos^4 (\alpha) - sin^4 (\alpha) =(cos^2(\alpha) + sin^2(\alpha))(cos^2(\alpha)-sin^2(\alpha))=(cos^2(\alpha)-sin^2(\alpha))$

That's correct til here.

As $cos(2\alpha)=cos^2(\alpha)-sin^2(\alpha)$ the numerator is zero, and the denominator tends to zero... In this case the limit will be zero. But as x is not four, but tends to four, I'm not sure if the above trig. expressions can be used.

0/0 is an undetermined limit !!!!!!

Anyway, then you can notice that $\cos^4\alpha-\sin^4\alpha=\cos^2\alpha-\sin^2\alpha=\cos(2\alpha)$

So if you let $f(x)=\cos^x\alpha-\sin^x\alpha$ , your limit is actually :
$\lim_{x\to 4} \frac{f(x)-f(4)}{x-4}$ , which is (the difference quotient of the function f) equal to $f'(4)$

**pankaj** · June 9th 2009, 10:14 AM

$\lim_{x\rightarrow 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}$

$=\lim_{x\rightarrow 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos^4\alpha+\sin^4\alpha}{x - 4}$

$=\lim_{x\rightarrow 4} \frac{(\cos \alpha)^x -\cos^4\alpha- ((\sin \alpha)^x -\sin^4\alpha)}{x - 4}$

$=\lim_{x\rightarrow 4} \frac{\cos^4\alpha((\cos \alpha)^{x-4} -1)-\sin^4\alpha ((\sin \alpha)^{x-4} -1)}{x - 4}$

$=\cos^4\alpha.ln(\cos\alpha)-\sin^4\alpha.ln(\sin\alpha)$

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