Question

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

Answer 1

Hint: Indeterminate form is any form that can’t be evaluated, like ${\displaystyle \frac{0}{0}},{\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}},0.\mathrm{\infty },\mathrm{\infty }.0$ 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.
${\displaystyle \frac{0}{0}},{\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}},0.\mathrm{\infty },0^{{}^{\text{0}}},{\mathrm{\infty }}^{{}^{\text{0}}},1^{\mathrm{\infty }},\mathrm{\infty }-\mathrm{\infty }$ are all called indeterminate forms, because they can’t be evaluated to a number.

We will check one by one:
(a) $\underset{x\to a}{lim}{\displaystyle \frac{x^3-a^3}{x-a}}$
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 :
$\underset{x\to a}{lim}{\displaystyle \frac{x^3-a^3}{x-a}}={\displaystyle \frac{a^3-a^3}{a-a}}={\displaystyle \frac{0}{0}}$
Since ${\displaystyle \frac{0}{0}}$ falls under the category of an indeterminate form, this limit is in the indeterminate
form.
(b) $\underset{x\to 0}{lim}\sin x\cos \text{ec }x$
Similarly, let’s substitute for $x$ or put $x=0$, and see what form we get.
Therefore, $\underset{x\to 0}{lim}\sin x\cos ecx=\sin (0)\cos ec(0)=0.{\displaystyle \frac{1}{0}}=0.\mathrm{\infty }$
Since the form we got here is $0.\mathrm{\infty }$, this limit is also an indeterminate form.
(c) $\underset{x\to 0}{lim}{x}^x$
Substituting for $x$ or putting $x=0$ in the limit, we get :
$\underset{x\to 0}{lim}{x}^x=0^0$
Since it does fall in the category of indeterminate forms, we can safely say that this limit also evaluates
to an indeterminate limit.
(d) $\underset{x\to 0}{lim}{\displaystyle \frac{0}{x}}$
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, ${\displaystyle \frac{0}{x}}=0$
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.