Question

Find the value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 3x}}{{\sin 5x}} = $

Answer 1

Hint: We can directly apply the limits. If the limit becomes $\dfrac{0}{0}$ , we can apply L’ Hospital’s rule which is given by $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ . We can find the derivatives of the functions separately. Then we can substitute in the equation. Then we can apply the limits to get the value of the required limit.

Complete step-by-step answer:
We need to find the value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 3x}}{{\sin 5x}}$
Let $I = \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 3x}}{{\sin 5x}}$
We can directly apply the limits. For that we can substitute $x = 0$ .
 $ \Rightarrow I = \dfrac{{\tan \left( {3 \times 0} \right)}}{{\sin \left( {5 \times 0} \right)}}$
On further simplification, we get
 $ \Rightarrow I = \dfrac{{\tan 0}}{{\sin 0}}$
We know that, $\tan 0 = 0$ and $\sin 0 = 0$ . On applying this relation, we get,
 $ \Rightarrow I = \dfrac{0}{0}$
So, we can apply L’ Hospital’s rule. According to this rule, if $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}$ ,then $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ .
Here $f\left( x \right) = \tan 3x$ and $g\left( x \right) = \sin 5x$
Now we can find $f'\left( x \right)$ and $g'\left( x \right)$
 $f'\left( x \right) = \dfrac{d}{{dx}}\tan 3x$
We know that, $\dfrac{d}{{dx}}\tan x = {\sec ^2}x$ and by applying chain rule of differentiation, we get,
 $f'\left( x \right) = 3 \times {\sec ^2}3x$
Now $g'\left( x \right)$ is given by,
 $g'\left( x \right) = \dfrac{d}{{dx}}\sin 5x$
We know that, $\dfrac{d}{{dx}}\sin x = \cos x$ and by applying chain rule of differentiation, we get,
 $g'\left( x \right) = 5 \times \cos 5x$
By L’ Hospital’s Rule, $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ , we can write the given limit as,
 $ \Rightarrow I = \mathop {\lim }\limits_{x \to 0} \dfrac{{3{{\sec }^2}3x}}{{5\cos 5x}}$
On applying the limits, we get,
 $ \Rightarrow I = \dfrac{{3{{\sec }^2}0}}{{5\cos 0}}$
We know that $\cos 0 = 1$ and $\sec 0 = \dfrac{1}{{\cos 0}} = 1$ . On substituting these values, we get,
 $ \Rightarrow I = \dfrac{{3 \times {1^2}}}{{5 \times 1}}$
On simplification, we get,
 $ \Rightarrow I = \dfrac{3}{5}$
So, the required limit is $\dfrac{3}{5}$
Therefore, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 3x}}{{\sin 5x}} = \dfrac{3}{5}$.

Note: Limit of a function will give the value at which the function tends to when x tends to a point. When we get an expression to find the limit, we can directly apply the limits to check whether it is defined. Then we can use different methods to simplify the expression and then apply the limits to get the required value of limit. We can use only the L’ Hospital’s rule $\mathop {\lim }\limits_{x \to \infty } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ , only when the limit tends to $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ . While taking the derivatives, we must use the chain rule. We must also know the derivatives of standard trigonometric functions.