Answer

Verified

452.4k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How many crores make 10 million class 7 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE