Question

Which of the following limits is in the indeterminate form?
(a) $\displaystyle \lim_{x \to \infty }{{\left( \dfrac{1}{7} \right)}^{x}}$
(b) $\displaystyle \lim_{x \to \infty }{{1}^{x}}$
(c) $\displaystyle \lim_{x \to \infty }{{5}^{x}}$
(d) $\displaystyle \lim_{x \to \infty }{{\left( 1+\dfrac{1}{x} \right)}^{x}}$

Answer 1

In this question, we have to find whether the given options are of the Indeterminate forms or not. Thus, we know that an indeterminate form means when the limit of two functions are not determined solely from the limit of the individual function. Thus, in this problem, we will solve all the four parts by using logarithm function and the basic mathematical rule, to get the solution.

Complete step by step answer:
As, we know that an indeterminate form means when the limit of two functions are not determined solely from the limit of the individual function. The forms $\dfrac{0}{0},\dfrac{\infty }{\infty },0.\infty ,{{0}^{{}^\circ }},{{\infty }^{{}^\circ }},{{1}^{\infty }},\infty -\infty$ are all called indeterminate forms.
Thus, let us solve all the four parts of the given problem.
(a) $\displaystyle \lim_{x \to \infty }{{\left( \dfrac{1}{7} \right)}^{x}}$
Let us put the log function in the above limit, we get
$\Rightarrow \log \left( \displaystyle \lim_{x \to \infty }{{\left( \dfrac{1}{7} \right)}^{x}} \right)$
Now, we will apply the log-limit formula $\log \left( \lim x \right)=\lim \left( \log x \right)$ in the above equation, we get
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( \log {{\left( \dfrac{1}{7} \right)}^{x}} \right)$
So, on solving the log function, we get
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( x\log \left( \dfrac{1}{7} \right) \right)$
Now, we will apply the log formula $\log \left( \dfrac{a}{b} \right)=\log a-\log b$ in the above equation, we get
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( x\left( \log 1-\log 7 \right) \right)$
Now, we will apply the limit in place of x in the above equation, we get
$\Rightarrow \infty \left( \log 1-\log 7 \right)$
Also, log1=0, thus we get
$\Rightarrow \infty \left( 0-\log 7 \right)$
On further solving, we get
$\Rightarrow \infty \left( -\log 7 \right)$
Thus, we know that $\infty .-\log 7$ is not an indeterminate form, therefore $\displaystyle \lim_{x \to \infty }{{\left( \dfrac{1}{7} \right)}^{x}}$ is not an indeterminate form.
(b) $\displaystyle \lim_{x \to \infty }{{1}^{x}}$
Let us put the log function in the above limit, we get
$\Rightarrow \log \left( \displaystyle \lim_{x \to \infty }{{1}^{x}} \right)$
Now, we will apply the log-limit formula $\log \left( \lim x \right)=\lim \left( \log x \right)$ in the above equation, we get
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( \log {{1}^{x}} \right)$
So, on solving the log function, we get
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( x\log 1 \right)$
Now, we will apply the limit in place of x in the above equation, we get
$\Rightarrow \infty .\log 1$
Also, log1=0, thus we get
$\Rightarrow \infty .0$
Thus, we know that $\infty .0$ is an indeterminate form, therefore $\displaystyle \lim_{x \to \infty }{{1}^{x}}$ is an indeterminate form.
(c) $\displaystyle \lim_{x \to \infty }{{5}^{x}}$
Let us put the log function in the above limit, we get
$\Rightarrow \log \left( \displaystyle \lim_{x \to \infty }{{5}^{x}} \right)$
Now, we will apply the log-limit formula $\log \left( \lim x \right)=\lim \left( \log x \right)$ in the above equation, we get
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( \log {{5}^{x}} \right)$
So, on solving the log function, we get
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( x\log 5 \right)$
Now, we will apply the limit in place of x in the above equation, we get
$\Rightarrow \infty .\log 5$
Thus, we did not get any form of indeterminate form, therefore $\displaystyle \lim_{x \to \infty }{{5}^{x}}$ is not an indeterminate form.
(d) $\displaystyle \lim_{x \to \infty }{{\left( 1+\dfrac{1}{x} \right)}^{x}}$
So, let us first substitute $x=\infty$ in the above equation, we get
$\Rightarrow {{\left( 1+\dfrac{1}{\infty } \right)}^{\infty }}$
As we know that, $\dfrac{1}{\infty }=0$ , therefore we get
$\Rightarrow {{\left( 1+0 \right)}^{\infty }}$
On further simplification, we get
$\Rightarrow {{\left( 1 \right)}^{\infty }}$
Thus, from part (b) we get that ${{1}^{\infty }}$ is an indeterminate form, therefore $\displaystyle \lim_{x \to \infty }{{\left( 1+\dfrac{1}{x} \right)}^{x}}$ is an indeterminate form.

So, the correct answer is “Option b and d”.

Note: While solving this problem, do mention all the steps properly to avoid confusion and mathematical errors. Do not confuse with $\displaystyle \lim_{x \to \infty }{{1}^{x}}$ and $\displaystyle \lim_{x \to \infty }{{5}^{x}}$ , here the base are different, thus the solution will be different. Also, mention the formula you are using to get the accurate solution.