Question

How do you find the limit of ${\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right)$ as ${\text{x}}$ approaches infinity using L’Hospital’s rule?

Answer 1

In this question, they asked us to find the limit of ${\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right)$ as ${\text{x}}$ approaching infinity using L’Hospital’s rule.
First we have to rewrite or change the expression in such a way that it becomes easier to use the rule of L’Hospital.
Then we have to differentiate the numerator and the denominator with respect to ${\text{x}}$ and simplify it, and then we will get the right answer.

Formula used: $\dfrac{{d\tan {\text{x}}}}{{dx}} = {\sec ^2}{\text{x}}$
Properties of trigonometric functions used:
$\tan {\text{x = }}\dfrac{{\sin {\text{x}}}}{{\cos {\text{x}}}}$
$\cos \left( 0 \right)$= $1$
Properties of limits used:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sin {\text{x}}}}{{\text{x}}} = 1$

Complete step-by-step solution:
We need to find the limit of ${\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right)$ as ${\text{x}}$ approaching infinity using L’Hospital’s rule.
Using L’Hospital’s rule we need to find$\mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right)$
Here$\mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right)$ has an indeterminate form $\infty$ .
First, we have to change or rewrite the expression in such a way that it becomes easier to use the rule of L’Hospital
Here we are changing ${\text{x}}$ into $\dfrac{{1{\text{ }}}}{{\dfrac{1}{{\text{x}}}}}$ for this purpose,
And we get,
$\Rightarrow {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = \dfrac{{{\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right)}}{{\dfrac{1}{{\text{x}}}}}$
Now we can apply the rule,
First, differentiating the expression with respect to ${\text{x}}$ , we get
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = \mathop {\lim }\limits_{x \to \infty } {\text{ }}\dfrac{{{{\sec }^2}\left( {\dfrac{9}{{\text{x}}}} \right)\left( { - \dfrac{9}{{{{\text{x}}^2}}}} \right)}}{{ - \dfrac{1}{{{{\text{x}}^2}}}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{ }}{\sec ^2}\left( {\dfrac{9}{{\text{x}}}} \right)\left( 9 \right)$
$\Rightarrow {\sec ^2}\left( {\dfrac{9}{\infty }} \right)\left( 9 \right)$
We know that ${\sec ^2}\left( 0 \right) = 1$, thus we get
$\Rightarrow {\sec ^2}\left( 0 \right)\left( 9 \right)$
$\Rightarrow 9$

Therefore $9$ is the required answer.

Note: In this question we have alternative method as follows
Alternative method:
We can find this without using the rule of L’Hospital’s
We have, ${\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right)$
We have to change the expression using the properties of trigonometric functions,
i.e. $\tan {\text{x = }}\dfrac{{\sin {\text{x}}}}{{\cos {\text{x}}}}$
we get, ${\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = {\text{x}}\dfrac{{\sin \left( {\dfrac{9}{{\text{x}}}} \right)}}{{\cos \left( {\dfrac{9}{{\text{x}}}} \right)}}$
Now changing the ${\text{x}}$ into $\dfrac{{1{\text{ }}}}{{{\text{ }}\dfrac{1}{{\text{x}}}{\text{ }}}}$,
$\Rightarrow {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = \dfrac{{\sin \left( {\dfrac{9}{{\text{x}}}} \right)}}{{\cos \left( {\dfrac{9}{{\text{x}}}} \right)\left( {\dfrac{1}{{\text{x}}}} \right)}}$
Now we have to multiply and divide $9$, we get
$\Rightarrow {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = 9 \times \dfrac{{\sin \left( {\dfrac{9}{{\text{x}}}} \right)}}{{\cos \left( {\dfrac{9}{{\text{x}}}} \right)\left( {\dfrac{1}{{\text{x}}}} \right)(9)}}$
$\Rightarrow {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = 9 \times \dfrac{{\sin \left( {\dfrac{9}{{\text{x}}}} \right)}}{{\left( {\dfrac{9}{{\text{x}}}} \right)}} \times \dfrac{1}{{\cos \left( {\dfrac{9}{{\text{x}}}} \right)}}$
Now applying the properties of limit,
i.e. $\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sin {\text{x}}}}{{\text{x}}} = 1$
So we get,
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = 9 \times \dfrac{{\sin \left( {\dfrac{9}{{\text{x}}}} \right)}}{{\left( {\dfrac{9}{{\text{x}}}} \right)}} \times \dfrac{1}{{\cos \left( {\dfrac{9}{{\text{x}}}} \right)}}$
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = 9(1)\dfrac{1}{{\cos \left( {\dfrac{9}{{\text{x}}}} \right)}}$
Apply limit ${\text{x}} \to \infty$ we get,
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = 9(1)\dfrac{1}{{\cos \left( {\dfrac{9}{\infty }} \right)}}$
That will make$\cos \left( {\dfrac{9}{\infty }} \right)$ to $\cos \left( 0 \right)$
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = 9(1)\dfrac{1}{{\cos \left( 0 \right)}}$
As we all know that $\cos \left( 0 \right)$= $1$ we get,
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = 9(1)(1)$
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{x tan}}\left( {\dfrac{9}{{\text{x}}}} \right) = 9$
And we got the required correct answer.