Question 190130
*[Tex \LARGE \lim_{\theta \to 0} \frac{\theta\tan(\theta)}{1-\cos(\theta)}] ... Start with the given limit



*[Tex \LARGE =\frac{0\tan(0)}{1-\cos(0)}] Evaluate the limit by plugging in 0 for each  *[Tex \LARGE \theta]



*[Tex \LARGE =\frac{0(0)}{1-1}] Evaluate the tangent of 0 to get 0.  Evaluate the cosine of 0 to get 1.



*[Tex \LARGE =\frac{0}{0}] Multiply and simplify



So *[Tex \LARGE \lim_{\theta \to 0} \frac{\theta\tan(\theta)}{1-\cos(\theta)}=\frac{0}{0}]



Since *[Tex \LARGE \frac{0}{0}] is an indeterminate form, this means that we must use L'Hospital's Rule to find the limit. Remember, <a href="http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule">L'Hospital's Rule</a> states that:


If functions f and g are in some <a href="http://en.wikipedia.org/wiki/Indeterminate_form">indeterminate form</a>, then


*[Tex \LARGE \lim_{x \to a}\frac{f(x)}{g(x)}=\lim_{x \to a}\frac{f'(x)}{g'(x)}] 



So let's use L'Hospital's Rule to find the limit:


*[Tex \LARGE \lim_{\theta \to 0} \frac{\theta\tan(\theta)}{1-\cos(\theta)}] ... Start with the original limit



*[Tex \LARGE \lim_{\theta \to 0} \frac{\tan(\theta)+\theta\sec^2(\theta)}{\sin(\theta)}] ... Derive the numerator and denominator separately




If you evaluate the limit for the last expression, you'll find that *[Tex \LARGE \lim_{\theta \to 0} \frac{\tan(\theta)+\theta\sec^2(\theta)}{\sin(\theta)}=\frac{0}{0}]




So we must use  L'Hospital's again:



*[Tex \LARGE \lim_{\theta \to 0} \frac{\tan(\theta)+\theta\sec^2(\theta)}{\sin(\theta)}] ... Start with the previous expression



*[Tex \LARGE \lim_{\theta \to 0} \frac{\sec^2(\theta)+\sec^2(\theta)+2\theta\sec(\theta)}{\cos(\theta)}] ... Derive the numerator and denominator separately



*[Tex \LARGE =\frac{2\sec^2(\theta)+2\theta\sec(\theta)}{\cos(\theta)}] ... Combine like terms



*[Tex \LARGE =\frac{2\sec^2(0)+2\cdot0\sec(0)}{\cos(0)}] ... Plug in *[Tex \LARGE \theta=0] to evaluate the limit



*[Tex \LARGE =\frac{2\sec^2(0)}{\cos(0)}] Multiply and simplify



*[Tex \LARGE =\frac{2(1)}{1}] Evaluate the cosine of 0 to get 1. Evaluate the secant squared of 0 to get 1/1 or just 1



*[Tex \LARGE =2] Multiply and reduce



So *[Tex \LARGE \lim_{\theta \to 0} \frac{\theta\tan(\theta)}{1-\cos(\theta)}=2]



Note: you can graph to visually confirm that the limit is 2.