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

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

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

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

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

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.

3. 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 $f'(4)$

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

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

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

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

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