# 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) $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{0}{x}$

Answer

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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.

},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.

Last updated date: 26th Sep 2023

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