Question

Evaluate \[ \mathop { \lim } \limits_{x \to 0} \dfrac{{ \ln (2 + x) + \ln (0.5)}}{x} \]

Answer 1

Hint: If we substitute the value of \[x = 0 \] in above we get in the form \[ \left( { \dfrac{0}{0}} \right) \] .
Meaning that we don’t know the value of this. So to solve this we use a rule called L’Hospital’s rule. If we use this method we will get the given function into a simplifying form. Then we will apply the limit and we don’t get an indeterminate form.

Complete step-by-step answer:
Given, \[ \mathop { \lim } \limits_{x \to 0} \dfrac{{ \ln (2 + x) + \ln (0.5)}}{x} \]
As \[x \] tends to zero we will get \[ \left( { \dfrac{0}{0}} \right) \] form.
That is, when we put \[x = 0 \] in the numerator its becomes,
 \[ = \ln (2) + \ln (0.5) \] ------- (1)
But, we know that \[ \ln (0.5) = \ln \left( { \dfrac{1}{2}} \right) \]
We know logarithmic identity, \[ \ln \left( { \dfrac{m}{n}} \right) = \ln m - \ln n \]
Here, \[m = 1 \] and \[n = 2 \]
Now equation (1) becomes,
 \[ = \ln (2) + \ln (1) - \ln (2) \]
Cancelling \[ \ln (2) \] we get
 \[ = 0 \] .
Denominator as \[x = 0 \] we get 0.
i.e., its \[ \left( { \dfrac{0}{0}} \right) \] form.
So applying L’Hospital’s rule. It states that if we get an indeterminate form after substituting \[x \] , we differentiate the numerator with respect to \[x \] and denominator with respect to \[x \] . And then applying \[ \lim \] . Again if we get indeterminate form we again apply L’Hospital’s rule. I.e., again differentiate numerator and denominator with respect to \[x \] . We proceed this rule until we get a definite value while applying \[ \lim \] .
Differentiating numerator and denominator with respect to \[x \] in \[ \mathop { \lim } \limits_{x \to 0} \dfrac{{ \ln (2 + x) + \ln (0.5)}}{x} \] .
We know that differentiation of \[ \log x \] with respect to \[x \] is \[ \dfrac{1}{x} \] ,
 Differentiation of constant is zero, since \[ \ln (0.5) \] ,
And differentiation of \[x \] with respect to \[x \] is 1, then
 \[ \Rightarrow \mathop { \lim } \limits_{x \to 0} \dfrac{{ \left( { \dfrac{1}{{2 + x}}} \right) + 0}}{1} \]
 \[ \Rightarrow \mathop { \lim } \limits_{x \to 0} \left( { \dfrac{1}{{2 + x}}} \right) \]
Applying \[ \lim \] as \[x \to 0 \] I.e., substituting \[x = 0 \] we get,
 \[ = \dfrac{1}{2} \]
Hence,
 \[ \mathop { \lim } \limits_{x \to 0} \dfrac{{ \ln (2 + x) + \ln (0.5)}}{x} = \dfrac{1}{2} \]
So, the correct answer is “ $\dfrac{1}{2} $ ”.

Note: We know that \[ \ln (0.5) = \ln (2) \] , that’s why we get \[ \left( { \dfrac{0}{0}} \right) \] form. If we get \[ \left( { \dfrac{ \infty }{ \infty }} \right) \] , \[ \left( { \dfrac{{ - \infty }}{ \infty }} \right) \] and any indeterminate form we use L’Hospital’s rule.
If the given equation can be reducible into simplest form such that it won’t take indeterminate form then we can solve without using L’Hospital’s rule. Indeterminate form means we don’t know the value of that form. Further in \[ \lim \] if we have \[x \to \infty \] that means we are not going to take \[x = \infty \] , we take values of \[x \] as much as bigger so that the answer will get closer and closer to 0.