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

How do you find the limit ln⁡(x2+1)x\dfrac{{\ln \left( {{x^2} + 1} \right)}}{x} as x→0x \to 0?

Explanation

Solution

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 ln⁡(x2+1)x\dfrac{{\ln \left( {{x^2} + 1} \right)}}{x} as x→0x \to 0.
This can be written as follows:-
We need to find the value of lim⁡x→0ln⁡(x2+1)x\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 lim⁡x→0ln⁡(x2+1)x\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {{x^2} + 1} \right)}}{x}, we will then obtain the following:-
⇒\RightarrowNumerator = 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 lim⁡x→0ln⁡(x2+1)x\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:-
⇒lim⁡x→0ln⁡(x2+1)x=lim⁡x→01x2+1×ddx(x2+1)1\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:-
⇒lim⁡x→0ln⁡(x2+1)x=lim⁡x→01x2+1×2x\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:-
⇒lim⁡x→0ln⁡(x2+1)x=lim⁡x→02xx2+1\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:-
⇒lim⁡x→0ln⁡(x2+1)x=01=0\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⁡(x2+1)\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)f(x) to be equal to ln⁡x\ln x and g(x)g(x) to be equal to x2+1{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 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty } form only.