I don't understand limit so much. For example I see $\lim_{x \to -3}$. And I always just put $-3$ everywhere I see $x$. I feel like I'm doing something wrong, but it seems correct all the time.

  • $\begingroup$ You cannot always replace the number. Consider the example $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1$ but you cannot replace $0$ $\endgroup$ – Dimitris Apr 29 '13 at 17:10
  • $\begingroup$ Three more examples: $\lim_{x\rightarrow 0} \frac{x^2}{x}=0$, $\lim_{x\rightarrow 0} \frac{x^2}{x^2}=1$, $\lim_{x\rightarrow 0} \frac{x}{x^2}$ undefined. $\endgroup$ – vadim123 Apr 29 '13 at 17:11
  • $\begingroup$ Noooo, I'm doing all the time wrong ? Math is more difficult than I thought ... :P $\endgroup$ – user75045 Apr 29 '13 at 17:12
  • $\begingroup$ If you take the function $f(x) = \begin{cases}0, & x=0 \\ 1, & x>0 \end{cases}$, then $\lim_{x \to 0} f(x) = 1$, but $f(0) = 0$. If a function is continuous, then you can evaluate at the limit point. $\endgroup$ – copper.hat Apr 29 '13 at 17:12
  • $\begingroup$ So there is problem if you get $0$ under ? $\endgroup$ – user75045 Apr 29 '13 at 17:13

Substitution "works" many times; it works but not always: $$\lim_{x\to a} f(x) = f(a)\quad \text{${\bf only}$ when $f(x)$ is defined and continuous at $a$}$$

and this is why understanding the limit of a function as the limiting value (or lack of one) when $x$ is approaching a particular value: getting very very near that value, is crucial. That is, $$\lim_{x \to a} f(x) \not\equiv f(a) \qquad\qquad\tag{"$\not \equiv$"$\;$ here meaning "not identically"}$$

E.g., your "method" won't work for $\;\;\lim_{x\to -3} \dfrac{x^2 - 9}{x + 3}\;\;$ straight off.

Immediate substitution $f(-3)$ evaluates to $\dfrac 00$ which is indeterminate: More work is required. Other examples are given in the comments.

When we seek to find the limit of a function $f(x)$ as $x \to a$, we are seeking the "limiting value" of $f(x)$ as the distance between $x$ and $a$ grows increasingly small. That value is not necessarily the value $f(a)$.

And understanding the "limit" as the "limiting value" or lack there of, of a function is crucial to understanding, e.g. that $\lim_{x \to +\infty} f(x)$ requires examining the behavior of $f(x)$ as $x$ gets arbitrarily (increasingly) large, where evaluating $f(\infty)$ to find the limit makes no sense and has no meaning.

  • $\begingroup$ so you don't replace -3 but, -2.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 $\endgroup$ – user75045 Apr 29 '13 at 17:14
  • $\begingroup$ This is so difficult ! $\endgroup$ – user75045 Apr 29 '13 at 17:15
  • $\begingroup$ So how I can do your problem ? You can make from $x^2-9$ to $(x-3)(x+3)$. Then you have only $(x-3)$. Can I replace that one ? $\endgroup$ – user75045 Apr 29 '13 at 17:17
  • $\begingroup$ Exactly, the example I give requires more work. When you cancel $(x + 3)$ you are left with only $(x - 3)$...and it is indeed safe then to substitute $x = -3$ to see that the limit is $-6$. My caution is that substitution won't always work. When you do substitute, and you are able to obtain a value, then you know that value is the limit. But it is best not to equate the concept of the limit of a function as x approaches a quantity, with it's value AT that quantity. The fact that it is often the case that the $\lim_{x\to a} f(x) = f(a)$ is handy to know, but will not always work. $\endgroup$ – Namaste Apr 29 '13 at 17:23
  • $\begingroup$ So yes, you can oftentimes replace $x$ with the value to which x is approaching to evaluate a limit. But keep in mind what a limit actually means. Go back to the definition of a limit...if necessary, to remind yourself about what is happening when taking a limit, versus merely evaluating a function at a given point. $\endgroup$ – Namaste Apr 29 '13 at 17:26

Limits are about the Journey, not the Destination

(I need to put this onto a T-shirt. :)

  • $\begingroup$ Would the downvoter care to explain the vote? To me, the slogan seems perfectly apt, and the image nicely illustrates (one possible instance of) the notion that a function's behavior when $x$ equals $a$ has no bearing on its behavior as $x$ approaches $a$ ... Confusion about this distinction is the essence of OP's question. $\endgroup$ – Blue Apr 30 '13 at 8:00
  • $\begingroup$ I agree with your comment above, @Blue so +1 from me! (+1 with or without the downvote...I like your post! And it IS a good slogan for the question at hand.) $\endgroup$ – Namaste Apr 30 '13 at 15:32
  • $\begingroup$ If this were on a T-shirt, I would wear it while teaching. $\endgroup$ – treble Apr 30 '13 at 15:58

Let's go back to the $\epsilon,\delta$ definition of a limit.

Let a function $f$ be defined on an open interval containing the real number $c$. -This mean we aren't going to do silly things like take the square root of negative numbers etc. IN some interval around $c$.

Let $L$ be a real number -Self explanatory.

Then $\lim_{x\rightarrow c}f(x)=L$ if and only if -This is about to be an equivalent statement to what comes next.

For any real number $\epsilon>0$ -So we're allowing $\epsilon$ to be as absolutely small as we like, as long as it's still positive.

There exists a real number $\delta$ with the following property -Ready for this?

For all $x$ if $0<|x-c|<\delta$ then $|f(x)-L|<\epsilon$ -This means that if $|x-c|$ (think of this as the distance between $x$ and $c$) is less than our number $\delta$ then $f(x)$ is arbitrarily close to the real number $L$.

I'm not sure if this was very helpful but let me end by suggesting that you look up visual representations of the above as well as visuals of where a limit does not exist.


One of the most important limits is the definition of the derivative $$ \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} $$ and you cannot substitute $h=0$ in that fraction.


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