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