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If $\ f:\mathbb R\rightarrow \mathbb R\ $ where $f(x)=\frac{1}{x^2}$ then $$\lim_{x\to0}f(x)=?$$

Which of the following option is most correct among these

(a)$\infty$

(b) limit does not exists

Solution

According to me$$\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=\infty$$

but the limit is not finite so we can say that the limit does not exists

I am confused to choose the option among (a) & (b) in single choice type question

Can anyone tell me ?

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  • $\begingroup$ Note that $f$ as given is not a function, as it is not defined at $0\in\Bbb{R}$. $\endgroup$ – Servaes Jun 4 '16 at 11:18
  • $\begingroup$ Negative part isn't in the domain of $x^2$ so b is more appropriate $\endgroup$ – Archis Welankar Jun 8 '16 at 12:23
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The limit in fact does not exist, as $\infty$ is not a value but more a kind of expression. But I think everyone will know that by saying $\lim_{x\to 0}f(x)=\infty$ you mean $f(x) \to \infty$ for $x\to 0$.

When beginning to work with limits it is didactically-wise more meaningful to use the the latter, but later on it doesn't matter.

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The limit does not exist in the sense that there is no real number $L$ for which the formal definition of $f(x) \to L$ as $x \to a$ (with $a=0$ in your case) is satisfied; you have probably seen this $\varepsilon \; \delta$-definition and if a function satisfies it, the following notation is used: $$\lim_{x \to a} f(x) = L$$

Usually however, the concept of limits is generalized to capture some other kinds of (limit) behavior. One of those is $f(x) \to +\infty$ as $x \to a$. Note that this has a different formal definition. Your example satisfies this definition, so if you have seen this generalization: it will be the answer and the following notation is introduced: $$\lim_{x \to a} f(x) = +\infty$$

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The limit is correct, the definition says that a limit of a function $f$ does not exists if and only if doesn't exists the limite finite in $\mathbb R$ and the limit $\pm\infty$. So $lim_{x\to x_0 }f=l$ exists if $l \in \mathbb R, $ or $l= \pm\infty$. Hope this help you.

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  • $\begingroup$ but which option will be correct to choose among them? c an you tell me $\endgroup$ – Girish Kumar Chandora Jun 4 '16 at 9:13
  • $\begingroup$ According to the classic definition the correct option is (a). $\endgroup$ – Marco Lecci Jun 4 '16 at 9:15
  • $\begingroup$ whats wrong with the option b if w choose ti instead of a $\endgroup$ – Girish Kumar Chandora Jun 4 '16 at 10:01
  • $\begingroup$ I think it is wrong because i'm appealling to the definition that i wrote. $\endgroup$ – Marco Lecci Jun 4 '16 at 10:04
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Technically, (b) is the most correct. However, we've invented a shorthand for the times when the function grows larger and larger without coming back down as $x \to 0$. That shorthand is written $\lim_{x \to 0}f(x) = \infty$. Keep in mind, though, that it is a shorthand, and it means that for any $N>0$, we may find a $\delta > 0$ such that $|f(x)| > N$ for all $0<|x|<\delta$. That is a fundamentally different statement from $\lim_{x \to 0}f(x) = L$ for any finite $L$.

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  • $\begingroup$ Shouldn't it be $x \geq \delta$ $\endgroup$ – Gathdi Jun 4 '16 at 9:51
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    $\begingroup$ @Gathdi No. This is the limit as $x \to 0$, so we are setting an upper bound for $x$ that lets us say that $f(x)$ is large enough. $\endgroup$ – Arthur Jun 4 '16 at 9:53
  • $\begingroup$ @Gathdi There should, however, have been absolute value signs on my $x$. That has been fixed. $\endgroup$ – Arthur Jun 4 '16 at 9:54
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Usually we say that a limit exists if it is equal to a number. This is the convention used in, for example, Stewart's Calculus book. Infinity ($\infty$) is not a number and so if a limit is equal to $\infty$, then it does not exist. With this convention the answer is (a) and (b).

Now, this is the standard convention, but there are probably books/notes/teachers who use other conventions. So to accurately answer your question, you need to ask your teacher or consult your textbook/notes.

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