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I have struggled with understanding indeterminate forms for quite sometime now. In this post, I put forward my understanding and hope to see it expand( or get corrected ).

Some limits of functions are certain at first observation, others are not. These "others", at a first look, take indeterminate forms as the independent variable gets sufficiently close to a point. By first look, I mean that an indeterminate form may crop up when we make a direct substitution in potential functions. We see that their limits take the following forms.

$0^0$; $1^\infty$; $0.\infty$; $\frac{\infty}{\infty}$; $\infty - \infty$; $\frac{0}{0}$

The term indeterminate, if I understand well, means that we are still not sure what the original value is( original limit as per the context of this post ). The original limit could be any real number, infinity or undefined( does not exist ). To properly convey this concept, I have constructed some examples. I shall deal with the above cases.

1) $0^0$

$\lim_{x\rightarrow 0} x^0 = 1$; $\lim_{x\rightarrow 0} 0^{|x|} =0$.

At first look, the above two limits take indeterminate forms. Their limits, however, are different. This tells us that an indeterminate form can take multiple real values.

2) $\frac{0}{0}$

$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$; $\lim_{x\rightarrow 0}\frac{|x|}{x}$ is undefined.

While both the limits take indeterminate forms at first look, the first takes a real value and the second's limit is undefined.

3) $1^\infty$

$\lim_{x\rightarrow 1} x^{\frac{1}{x-1}}=e$; $\lim_{x\rightarrow 1} x^{\frac{1}{|x-1|}}$ does not exist.

4) $\frac{\infty}{\infty}$

$\lim_{x\rightarrow 0} \frac{\frac{1}{x}}{\frac{1}{x}}=1$; $\lim_{x\rightarrow 0} \frac{\frac{1}{|x|}}{\frac{1}{x}}$ does not exist.

5) $0.\infty$

$\lim_{x\rightarrow 0} x.\frac{1}{x}=1$; $\lim_{x\rightarrow 0} \sin(x).\frac{1}{|x|}$ does not exist.

6) $\infty - \infty$

$\lim_{x\rightarrow 0} \frac{1}{x}-\frac{1}{x}=0$; $\lim_{x\rightarrow 0} \frac{1}{x}-\frac{1}{|x|}$ does not exist.

From the above, we see that indeterminate forms can take several values. Often while computing limits, we run into these forms. We then perform algebraic manipulations to arrive at the original limit( could exist or could not exist ). I haven't stated an Infinite limit in my examples. The following is one.

$\lim_{x\rightarrow 0} \frac{x}{x^2}$ is an infinite limit that, at first look, takes the indeterminate form $\frac{0}{0}$.

I end this post with a question. Are there other indeterminate forms?

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    $\begingroup$ Taking logs on $1^{\infty}$ and $\infty^{0}$ reduces them to $0\cdot\infty$ and essentially these forms have to be dealt with in this manner by taking logs so from practical point of view of evaluation of limits they can be considered as equivalent. Btw what you have written about indeterminate forms with examples is very good. Especially when you arrived at this understanding yourself. +1 $\endgroup$ – Paramanand Singh Jul 29 '17 at 6:17
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I think you just missed one indeterminate form which is $\infty^0$, an example of this type of indeterminate form could be:

$lim_{n \to 0^+} (\frac{1}{n})^n$=1.

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  • $\begingroup$ I have a concern here. For $n$ negative, say $n=\frac{-1}{2(213465975135915754678....)}$, which can be written as $(\frac{1}{2}).(\frac{-1}{213465975135915754678....})$, the limit will not exist( by knowing $a^{mn } = (a^m)^n$ and analyzing ) as square root of a negative number is not real. $\endgroup$ – R004 Jul 29 '17 at 4:04
  • $\begingroup$ I can take $lim_{x\rightarrow 0} (\frac{1}{|x|})^x$. Now the limit of this is 1. We can test our results by plotting graphs. $\endgroup$ – R004 Jul 29 '17 at 4:13
  • $\begingroup$ When you say $n\rightarrow 0$, you know for a fact that $n$ takes infinitely many values close to zero. Here, it's not in your hand to take even or odd values, don't you think so? $\endgroup$ – R004 Jul 29 '17 at 5:19
  • $\begingroup$ I have plotted it in Wolfram to see that for negative $n$, the imaginary part is not zero. As I mentioned earlier, the limit is not real for $n<0$. $\endgroup$ – R004 Jul 29 '17 at 5:26
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    $\begingroup$ You're right R004. Since this is just an example we consider $n \to 0^+$ or the example you wrote with the absolute value. Regards. $\endgroup$ – cristiam.chica Jul 29 '17 at 12:06

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