Limit
- Thread starter alexmay
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Hello, and welcome to FMH! 
Assuming you have not been exposed to L'Hôpital's Rule, I would instead make use of:
[MATH]\lim_{x\to0}\frac{\sin(x)}{x}=1[/MATH]
And by extension:
[MATH]\lim_{x\to0}\frac{\sin(ax)}{ax}=1[/MATH] where \(a\ne0\). Suppose we write:
[MATH]L=\lim_{\theta\to0}\frac{\sin(9\theta)}{\sin(11\theta)}=\lim_{\theta\to0}\frac{9\theta\dfrac{\sin(9\theta)}{9\theta}}{11\theta\dfrac{\sin(11\theta)}{11\theta}}[/MATH]
Can you proceed?
Assuming you have not been exposed to L'Hôpital's Rule, I would instead make use of:
[MATH]\lim_{x\to0}\frac{\sin(x)}{x}=1[/MATH]
And by extension:
[MATH]\lim_{x\to0}\frac{\sin(ax)}{ax}=1[/MATH] where \(a\ne0\). Suppose we write:
[MATH]L=\lim_{\theta\to0}\frac{\sin(9\theta)}{\sin(11\theta)}=\lim_{\theta\to0}\frac{9\theta\dfrac{\sin(9\theta)}{9\theta}}{11\theta\dfrac{\sin(11\theta)}{11\theta}}[/MATH]
Can you proceed?
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I am multiplying the numerator by 1 (so it retains the same value in the form:
[MATH]1=\frac{9\theta}{9\theta}[/MATH]
And likewise I am multiplying the denominator by:
[MATH]1=\frac{11\theta}{11\theta}[/MATH]
So, the value of the limit will be the same as the original. And then we can write:
[MATH]L=\frac{9}{11}\lim_{\theta\to0}\frac{\dfrac{\sin(9\theta)}{9\theta}}{\dfrac{\sin(11\theta)}{11\theta}}[/MATH]
Next, we can consider the following rule for limits:
[MATH]\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to c} f(x)}{\lim\limits_{x\to c} g(x)}[/MATH]
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Next, I would write:
[MATH]L=\frac{9}{11}\frac{\lim\limits_{\theta\to0}\dfrac{\sin(9\theta)}{9\theta}}{\lim\limits_{\theta\to0}\dfrac{\sin(11\theta)}{11\theta}}[/MATH]