Which of the following limit is not in the indeterminate form?
(a) $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{3}}-{{a}^{3}}}{x-a}$
(b) $\underset{x\to 0}{\mathop{\lim }}\,\sin x\text{ }\cos \text{ec }x$
(c) $\underset{x\to 0}{\mathop{\lim }}\,{{x}^{x}}$
(d) $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{0}{x}$

Question

Which of the following limit is not in the indeterminate form?
(a) $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{3}}-{{a}^{3}}}{x-a}$
(b) $\underset{x\to 0}{\mathop{\lim }}\,\sin x\text{ }\cos \text{ec }x$
(c) $\underset{x\to 0}{\mathop{\lim }}\,{{x}^{x}}$
(d) $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{0}{x}$

Vedantu Content Team · Accepted Answer

Hint: Indeterminate form is any form that can’t be evaluated, like $\dfrac{0}{0},\dfrac{\infty }{\infty
},0.\infty ,\infty .0$ etc.
Now, firstly we have to know what indeterminate form;
After putting a limit, we see if the equation forms any of the forms mentioned below.
$\dfrac{0}{0},\dfrac{\infty }{\infty },0.\infty ,{{0}^{{}^\0}},{{\infty }^{{}^\0 }},{{1}^{\infty }},\infty -
\infty $ are all called indeterminate forms, because they can’t be evaluated to a number.

We will check one by one:
(a) $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{3}}-{{a}^{3}}}{x-a}$
Now, to see what form we get on substituting the limiting value of $x$, let’s actually do so. So, we’re
going to put $x=a$ in the limit given to us above. Doing so, we get :
$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{3}}-{{a}^{3}}}{x-a}=\dfrac{{{a}^{3}}-{{a}^{3}}}{a-
a}=\dfrac{0}{0}$
Since $\dfrac{0}{0}$ falls under the category of an indeterminate form, this limit is in the indeterminate
form.
(b) $\underset{x\to 0}{\mathop{\lim }}\,\sin x\text{ }\cos \text{ec }x$
Similarly, let’s substitute for $x$ or put $x=0$, and see what form we get.
Therefore, $\underset{x\to 0}{\mathop{\lim }}\,\sin x\cos ecx=\sin (0)\cos ec(0)=0.\dfrac{1}{0}=0.\infty $
Since the form we got here is $0.\infty $, this limit is also an indeterminate form.
(c) $\underset{x\to 0}{\mathop{\lim }}\,{{x}^{x}}$
Substituting for $x$ or putting $x=0$ in the limit, we get :
$\underset{x\to 0}{\mathop{\lim }}\,{{x}^{x}}={{0}^{0}}$
Since it does fall in the category of indeterminate forms, we can safely say that this limit also evaluates
to an indeterminate limit.
(d) $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{0}{x}$
Now, in this question, we can easily just substitute for $x$ and say that the limit gives an indeterminate
form, but what we also have to see is that the expression can be further simplified without putting the
value of $x$, and it will remain the same no matter what value of $x$ you put. Since, $\dfrac{0}{x}=0$
for any value of $x$, this limit will always be equal to $0$, and $0$ is not an indeterminate form.
Hence, this is the limit which doesn’t give an indeterminate form.
Therefore, option (d) is correct.
Note: Make sure to be well versed with the indeterminate forms, and if you get confused before
substituting for $x$, then try substituting for a different value of $x$ and see if the limit’s value changes.
If it doesn’t, then the limit is defined. If it does, and gives a defined value on changing the value of $x$
substituted, then the limit actually is undefined at the limiting value given to us, in the question. For
example, in option d, no matter what value of $x$ we used, the limit would always give us $0$,
however, in option a, if we put $x=2a$ for example, then we’d definitely get a defined limit. Thus, the
limit for option a and the other options changed when we changed the limiting value of $x$, and hence,
they were essentially indeterminant at the limiting value given to us.

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