Question

Find the limit of $f\left( x \right) = \dfrac{{\sec x - 1}}{{{x^2}}}$ as $x$ approaches 0?

Answer 1

As we have to find the limit of the function as $x$ approaches 0. Since both numerator and denominator give 0 for $x = 0$. So, apply the L-Hospital rule to remove the redundancy. After that use the formula $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1$ to get the limit of the function.

Complete step by step answer:
The given function is $f\left( x \right) = \dfrac{{\sec x - 1}}{{{x^2}}}$.
We have to find the limit of the function as $x$ approaches 0.
Since the function will give $\dfrac{0}{0}$ for $x = 0$, we will apply L’Hospital’s rule. According to this,
$\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
Now, according to L’Hospital’s rule $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$, but this relation has some conditions. The criteria are for the independent limits of f(x) and g(x) functions to be equal to zero or infinity that is $\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} g\left( x \right) = 0, \pm \infty$ and the derivative in the denominator should not be equal to zero that is $g'\left( x \right) \ne 0$ and the limit must exist.
So, applying L’Hospital’s rule in the function, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sec x - 1}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sec x\tan x}}{{2x}}$
As $\sec x$ will give 1 when $x$ approaches 0. So,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sec x - 1}}{{{x^2}}} = \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \sec x \times \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x}$
Use the formula, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1$ in the above limit,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sec x - 1}}{{{x^2}}} = \dfrac{1}{2} \times \sec 0 \times 1$
Simplify the terms,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sec x - 1}}{{{x^2}}} = \dfrac{1}{2}$
Hence, the limit of the function is $\dfrac{1}{2}$.

Note: A function derivative is a metric for figuring out the instantaneous change in a function. A differential equation is considered an equation with one or more derivatives. So in this matter, we use the rule of L'Hospital, the rule of L'Hospital allows one to find the boundaries of an indifferential/indeterminate equation. This law transforms an indeterminate equation to a form comprising both numerator and denominator differentials that can be conveniently checked by translating it to a boundary. Thus, we can figure out the specified limit by using the L'Hospital law.