Question

How do you find the limit of $\dfrac{{{{\tan }^2}x}}{x}$ as $x \to 0?$

Answer 1

Hint : This question involves the operation of addition/ subtraction/ multiplication/ division. We need to know how to expand the square terms. We need to know the basic trigonometric conditions. We need to know how to apply the limit with the given equation. We need to know the trigonometric table values to make the easy calculation.

Complete step-by-step answer :
The given equation in the question is shown below,
$\dfrac{{{{\tan }^2}x}}{x}$ as $x \to 0?$
The above equation can also be written as,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^2}x}}{x} = ?$ $\to \left( 1 \right)$
To solve the above equation, we have to solve the above-mentioned term as follows,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^2}x}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x}.\tan x \to \left( 2 \right)$
We know that,
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
So, the equation $\left( 2 \right)$ can be written as,

$$\left( 2 \right) \to \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^2}x}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} \cdot \tan x \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} \cdot \tan x = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{x} \cdot \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{{\sin x}}{{\cos x}} \\\$$

The above equation can also be written as,
$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x} \cdot \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{{\sin x}}{{\cos x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{1}{{\cos x}}$
We know that,
$\cos x \cdot \cos x = {\cos ^2}x$
So, we get
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{1}{{\cos x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \sin x \cdot \dfrac{1}{{{{\cos }^2}x}}$
Let’s apply the limit in the above equation we get,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{1}{{\cos x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \sin x \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{{\cos }^2}x}} \to \left( 3 \right)$
Here, we have $\mathop {\lim }\limits_{x \to 0}$ , so we would put zero for where we have $x$ .
For solving the equation $\left( 3 \right)$ , we have
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}$
We can’t apply the mentioned limit in the above equation, because if we apply the limit the denominator becomes zero. We know that the denominator would not be equal to zero.
Next, we have
$\mathop {\lim }\limits_{x \to 0} \sin x = \sin \left( 0 \right) = 0$ (By using trigonometric table values.)
Next, we have
$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}\left( 0 \right)}}$
We know that
$\cos 0 = 1$
So, we get
$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}\left( 0 \right)}} = \dfrac{1}{{{1^2}}} = 1$
By using these values the equation $\left( 3 \right)$ becomes,
$\left( 3 \right) \to \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{1}{{\cos x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \sin x \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{{\cos }^2}x}}$

$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{1}{{\cos x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot 0 \cdot 0 \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{1}{{\cos x}} = 0 \\\$$

(We know that if any term multiplies with $0$ the answer becomes $0$ .)
So, the final answer is,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^2}x}}{x} = 0$
So, the correct answer is “0”.

Note : Note that the denominator term would not be equal to zero and when any number is multiplied with zero the answer is always zero. $\mathop {\lim }\limits_{x \to 0}$ means we would apply zero for where we have the term $x$ . Remember the basic trigonometric conditions and trigonometric table values. This type of question involves basic arithmetic operations.