Question

The value of the limit when n tends to infinity of the expression ${\left( {1 + \dfrac{1}{n}} \right)^n}$ is
A) $e$
b) 0
C) 1
D) -1

Answer 1

To find the value of the limit first of all we have to let the given expression ${\left( {1 + \dfrac{1}{n}} \right)^n}$ is m (you can choose according to choice). Now, we will take $\ln$ on both sides of the obtained expression and then we will rearrange the expression with the help of the formula given below:
$\ln {a^b} = b\ln a$ ……………………………….(1)
After rearranging the expression we will put the limit as given in the question that n tends to infinity and on solving the expression we will see that after placing the limit of n the whole expression will become not defined so to get the value of the limit we will use the L’Hospital rule, Which is explained below.
L’Hospital rule says that the limit when we divide one function by another function is same after we take the derivative of each function.
$\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}$…………………………….(2)
Now, according to the rule given above, we will differentiate the whole expression, and then we will put the value of the limit n which is infinity as given in the question. After it, we will solve the $\ln$ to find the value.

Formula used:
$\dfrac{d}{{dx}}{a^n} = n{a^{n - 1}}...............................(3)$

Complete step by step answer:
Given,
N tends to infinity, and
Expression : ${\left( {1 + \dfrac{1}{n}} \right)^n}$
Step 1: First of all we will let that the given expression is equal to m hence,
$\Rightarrow m = {\left( {1 + \dfrac{1}{n}} \right)^n}$
Step 2: Now, we will take $\ln$ on both sides of the obtained expression.
$\Rightarrow \ln m = \ln {\left( {1 + \dfrac{1}{n}} \right)^n}$
Now, we will rearrange the expression with the help of the formula (1) as mentioned in the solution hint.
$\Rightarrow \ln m = n\ln \left( {1 + \dfrac{1}{n}} \right)$
Step 3: As given in the question now, we will put the value of the limit which is $n \to \infty$
Hence,
$\ln m = \mathop {\lim }\limits_{n \to \infty } n\ln \left( {1 + \dfrac{1}{n}} \right)$
On, rearranging the expression obtained just above,
$\Rightarrow \ln m = \dfrac{{\mathop {\lim }\limits_{n \to \infty } \ln \left( {1 + \dfrac{1}{n}} \right)}}{{{n^{ - 1}}}}$
Step 4: Now, we will substitute the value of limit in the equation obtained
$\Rightarrow \ln m = \dfrac{0}{0}$
Which is not defined.
Step 5: Hence, to obtain the value of limit we will use the L’Hospital rule …………………(2) as mention in the solution hint.
After applying L’Hospital rule,
$\Rightarrow \ln m = \dfrac{{\mathop {\lim }\limits_{n \to \infty } \dfrac{d}{{dx}}\left( {1 + \dfrac{1}{n}} \right)}}{{\dfrac{d}{{dx}}{n^{ - 1}}}}$
Step 6: Now, to solve the differentiation we will use the formula (3) as mentioned in the solution hint.
$\Rightarrow \ln m = \dfrac{{\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{{1 + \dfrac{1}{n}}}} \right)( - 1 \times {n^{ - 2}})}}{{ - 1 \times {n^{ - 2}}}}$
On solving the equation obtained just above,
$\Rightarrow \ln m = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{1 + \dfrac{1}{n}}}$
Step 7: Now, we will substitute the value of limit in the obtained expression.
$\Rightarrow \ln m = \dfrac{1}{{1 + \dfrac{1}{\infty }}}$
As we know $\infty = \dfrac{1}{0}$
$\Rightarrow \ln m = 1$…………………………………(4)
As, we know that $\ln e = 1$hence, on substituting the value of $\ln e$n equation (4)
$\Rightarrow \ln m = \ln e \\\ \Rightarrow m = e \\\$

Hence, with the help of the L’Hospital rule, we have obtained the value which is e. Therefore, our correct option is (A).

Note:
According to L’Hospital rule, the limit when we divide one function by another function is the same after we take the derivative of each function.
On solving the expression if the obtained values are like $\dfrac{0}{0}$or $\dfrac{1}{0}$after placing the value of the limit we can say that the limit of the function is not defined.
Try to make the expression in that form that after placing the limit according to the question limit does not become undefined.