Question

How do you find the limit of $x\ln (x)$ as $x$ approaches $0$?

Answer 1

Hint: We are asked to find the limit of the given function, we are given the value of $x$ as $0$ we will first try to put value of the variable directly into the function upon doing that we get that the $\ln x$ part of the function becomes undefined. So we will now solve our question by the use of L’Hospital Rule, we will first write our given function in the form that is suitable for the L’Hospital Rule that is forms that are $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form. Then solve by the rule and then substitute the value of $x$ in the obtained function to get the answer to the question.

Complete step by step solution:
The given question will be solved by using the L Hospital’s rule. We are using the L’Hospital rule because the given function becomes not defined at the given value of $x$ therefore we will now solve the given question by the L hospital rule we will first express our function in the $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form. We rewrite our function as:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{{\dfrac{1}{x}}}$.
Now to apply L Hospital rule we differentiate both the numerator and denominator from the above fraction then we get our value after differentiation as,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{{\dfrac{1}{x}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{x}}}{{ - \dfrac{1}{{{x^2}}}}}$.
Now we will apply the limit to the given function we get
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{x}}}{{ - \dfrac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to 0} \left( { - x} \right) = 0$.

Therefore, the limit of $x\ln (x)$ as $x$ approaches $0$ is $0$.

Note:
L'hospital theorem is only applicable when we get a function in an undefined form like $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$.