# What is the limit as t approaches 0 of (tan6t)/(sin2t)?

Oct 11, 2014

${\lim}_{t \to 0} \tan \frac{6 t}{\sin} \left(2 t\right) = 3$. We determine this by utilising L'hospital's Rule.

To paraphrase, L'Hospital's rule states that when given a limit of the form lim_(t→a)f(t)/g(t), where $f \left(a\right)$ and $g \left(a\right)$ are values that cause the limit to be indeterminate (most often, if both are 0, or some form of ∞), then as long as both functions are continuous and differentiable at and in the vicinity of $a ,$ one may state that

lim_(t→a)f(t)/g(t)=lim_(t→a)(f'(t))/(g'(t))

Or in words, the limit of the quotient of two functions is equal to the limit of the quotient of their derivatives.

In the example provided, we have $f \left(t\right) = \tan \left(6 t\right)$ and $g \left(t\right) = \sin \left(2 t\right)$. These functions are continuous and differentiable near $t = 0 , \tan \left(0\right) = 0 \mathmr{and} \sin \left(0\right) = 0$. Thus, our initial f(a)/g(a)=0/0=?.

Therefore, we should make use of L'Hospital's Rule. $\frac{d}{\mathrm{dt}} \tan \left(6 t\right) = 6 {\sec}^{2} \left(6 t\right) , \frac{d}{\mathrm{dt}} \sin \left(2 t\right) = 2 \cos \left(2 t\right)$. Thus...

${\lim}_{t \to 0} \tan \frac{6 t}{\sin} \left(2 t\right) = {\lim}_{t \to 0} \frac{6 {\sec}^{2} \left(6 t\right)}{2 \cos \left(2 t\right)} = \frac{6 {\sec}^{2} \left(0\right)}{2 \cos \left(0\right)} = \frac{6}{2 \cdot {\cos}^{2} \left(0\right) \cdot \cos \left(0\right)} = \frac{6}{2 \cdot 1 \cdot 1} = \frac{6}{2} = 3$

Jan 23, 2017

The Reqd. Lim.$= 3$.

#### Explanation:

We will find this Limit using the following Standard Results :

${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1 , {\lim}_{\theta \rightarrow 0} \tan \frac{\theta}{\theta} = 1$

Observe that, $\tan \frac{6 t}{\sin} \left(2 t\right) = \frac{\tan \frac{6 t}{6 t}}{\sin \frac{2 t}{2 t}}$$\frac{6 t}{2 t} = 3 \frac{\tan \frac{6 t}{6 t}}{\sin \frac{2 t}{2 t}}$

Here, $t \rightarrow 0 \Rightarrow \left(6 t\right) \rightarrow 0 \Rightarrow {\lim}_{t \rightarrow 0} \tan \frac{6 t}{6 t} = 1$

Similarly, ${\lim}_{t \rightarrow 0} \sin \frac{2 t}{2 t} = 1$

Therefore, the Reqd. Lim.$= 3 \left\{\frac{1}{1}\right\} = 3$.