Question

How do you find the limit $\dfrac{{\ln \left( {{x^2} + 1} \right)}}{x}$ as $x \to 0$?

Answer 1

Hint: We will first put in x = 0 and observe that we have obtained the 0/0 form. Then, we will use the L Hospital rule which will require us to differentiate both numerator and denominator separately. After that, put in x = 0.

Complete step by step solution:
We are given that we are required to find the limit of $\dfrac{{\ln \left( {{x^2} + 1} \right)}}{x}$ as $x \to 0$.
This can be written as follows:-
We need to find the value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x}$.
Now, if we put in x = 0 in the numerator of $\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x}$, we will then obtain the following:-
$ \Rightarrow $Numerator = ln (0 + 1) = ln 1 = 0
Therefore, the numerator of the given function becomes zero when we put in x = 0.
Now, if we put in x = 0 in the denominator of $\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x}$, we will then obtain the following:-
$ \Rightarrow $ Denominator = 0
Therefore, the denominator of the given function becomes zero when we put in x = 0.
Now, we have obtained the 0/0 form, therefore, using the L Hospital rule, we will find the derivatives of numerator and denominator and we will then obtain the following expression:-
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left\{ {\ln \left( {{x^2} + 1} \right)} \right\}}}{{\dfrac{d}{{dx}}\left( x \right)}}$
Simplifying the right hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{{x^2} + 1}} \times \dfrac{d}{{dx}}\left( {{x^2} + 1} \right)}}{1}$
Simplifying the right hand side of the above equation further, we will then obtain the following equation with us:-
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{x^2} + 1}} \times 2x$
We can right the above expression as follows:-
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{2x}}{{{x^2} + 1}}$
Now, putting in x = 0 in both numerator and denominator, we will then obtain the following expression with us:-
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x} = \dfrac{0}{1} = 0$

Hence, the required answer is 0.

Note: The students must note that while finding the derivative of $\ln \left( {{x^2} + 1} \right)$, we have used the chain rule which states that:-
$ \Rightarrow \dfrac{d}{{dx}}\left\{ {f\left( {g(x)} \right)} \right\} = f'(g(x)).g'(x)$
Here, we just had $f(x)$ to be equal to $\ln x$ and $g(x)$ to be equal to ${x^2} + 1$. Thus, we had the required derivative.
The students must also note that we pursue the L Hospital rule only when we have $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form only.