Question

How do you find the limit of $\dfrac{{\sec \left( {x - 1} \right)}}{{x\sec x}}$ as $x$ approaches $0$ ?

Answer 1

Hint : In order to determine the limit of the above function, you can see by putting $x = 0$ the result is in the indeterminate form i.e. $\dfrac{0}{0}\,or\,\dfrac{{ \pm \infty }}{{ \pm \infty }}$ . To remove the indeterminate form simplify the function into its simplest form then put the value of limit. If then also you get the same then check the left- hand limit and right-hand limit

Complete step by step solution:
We are given with the function $\dfrac{{\sec \left( {x - 1} \right)}}{{x\sec x}}$ as $x$ approaches $0$ which in limit form can be written as:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sec \left( {x - 1} \right)}}{{x\sec x}}$
Put the value of $x = 0$ and we get:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sec \left( {x - 1} \right)}}{{x\sec x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sec \left( {0 - 1} \right)}}{0}$ which is the indeterminate form.
Simplify the above function as $\sec x = \dfrac{1}{{\cos x}}$ :
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sec \left( {x - 1} \right)}}{{x\sec x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x}}{{x\cos \left( {x - 1} \right)}}$
Now put value of $x = 0$ as it cannot be further simplified and we get:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x}}{{x\cos \left( {x - 1} \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to 0} \cos x}}{{\mathop {\lim }\limits_{x \to 0} x\cos \left( {x - 1} \right)}} = \dfrac{1}{0}$ which we are gaining again the indeterminate form.
Now, lets check for left hand limit and right-hand limit:
For Left hand limit: $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\cos \left( {0 - h} \right)}}{{(0 - h)\cos \left( {0 - h - 1} \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to 0} \cosh }}{{\mathop {\lim }\limits_{x \to 0} - h\cos \left( {h + 1} \right)}} = - \dfrac{1}{0} = - \infty$
For right hand limit: $\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\cos \left( {0 + h} \right)}}{{(0 + h)\cos \left( {0 + h - 1} \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to 0} \cosh }}{{\mathop {\lim }\limits_{x \to 0} h\cos \left( {h - 1} \right)}} = \dfrac{1}{0} = \infty$
Since Left and right-hand limits are not the same.
Therefore, Limit for the function $\dfrac{{\sec \left( {x - 1} \right)}}{{x\sec x}}$ as $x$ approaches $0$ does not exist.

Note : Limit: You and your companions choose to meet at some spot outside. Is it essential that every one of your companions is living in a similar spot and stroll on a similar street to arrive at that place? Actually no, not generally. All companions come from various pieces of the city or nation to meet at that one single spot.

After putting the Limit the result should never in the indeterminate form $\dfrac{0}{0}\,or\,\dfrac{{ \pm \infty }}{{ \pm \infty }}$ . If it is contained, apply some operation to modify the result or use the L-Hospitals rule.