Question

Find $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}}.$

Answer 1

Hint : In this question, you have to evaluate the limit of a given function, and in order to proceed for the same, you have to simplify the given function in such a way that it will not give indeterminate form when limit is substituted. So, multiply and divide numerator and denominator both with their conjugates to simplify it.

Complete step-by-step answer :
In order to evaluate limit of the given function, that is $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}}$ we have to simplify the function such that we can put the limit without getting a indeterminate form, as you can see if we put the limit now, then we will get a indeterminate form as follows
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}} = \dfrac{{\sqrt {{0^2} + 1} - 1}}{{\sqrt {{0^2} + 16} - 4}} = \dfrac{{1 - 1}}{{4 - 4}} = \dfrac{0}{0}$
So here we are getting $\dfrac{0}{0}$ form,
We will multiply and divide the numerator and denominator with their conjugates to eliminate the $\dfrac{0}{0}$ form, we will get

$$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}} \\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}} \times \dfrac{{\sqrt {{x^2} + 1} + 1}}{{\sqrt {{x^2} + 1} + 1}} \times \dfrac{{\sqrt {{x^2} + 16} + 4}}{{\sqrt {{x^2} + 16} + 4}} \\\$$

Here we will use the algebraic identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ to simplify furthermore, we will get

$$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {{x^2} + 1} \right) - 1}}{{\left( {{x^2} + 16} \right) - 16}} \times \dfrac{{\sqrt {{x^2} + 16} + 4}}{{\sqrt {{x^2} + 1} + 1}} \\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{{x^2}}} \times \dfrac{{\sqrt {{x^2} + 16} + 4}}{{\sqrt {{x^2} + 1} + 1}} \\\ = \mathop {\lim }\limits_{x \to 0} 1 \times \dfrac{{\sqrt {{x^2} + 16} + 4}}{{\sqrt {{x^2} + 1} + 1}} \\\$$

Now, using distributive property of limit over division and putting limit in the simplified function, we will get
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 16} + 4}}{{\sqrt {{x^2} + 1} + 1}} = \dfrac{{\sqrt {{0^2} + 16} + 4}}{{\sqrt {{0^2} + 1} + 1}} = \dfrac{{\sqrt {16} + 4}}{{\sqrt 1 + 1}} = \dfrac{{4 + 4}}{{1 + 1}} = \dfrac{8}{2} = 4$
Therefore $4$ is the required answer of the given limit.
So, the correct answer is “4”.

Note : As you have seen that the given limit initially shown $\dfrac{0}{0}$ indeterminate form, so there is one more way to tackle or you may say solve $\dfrac{0}{0}$ indeterminate form limits, which is known as L’Hospital’s rule, in this rule you have to just differentiate the numerator and the denominator separately and then check the indeterminate form by putting the limit, if again found $\dfrac{0}{0}\;{\text{or}}\;\dfrac{\infty }{\infty }$ indeterminate form, then again apply L’Hospital’s rule. This rule is applicable in $\dfrac{0}{0}\;{\text{or}}\;\dfrac{\infty }{\infty }$ form.