0

In calculating the limit of $f(x)$ as $x$ approaches $a$, we never consider that $x=a$, but after simplifying why we put exactly $x=a$.

Please tell me.

Blue
  • 75,673
Danial Hussain
  • 9
  • If $a$ is in the domain of $f$, you can just substitute $x = a$. But often you want to compute the limit as $x$ approaches a point not in the domain, and that is where the limiting process is necessary. – DanLewis3264 Jul 23 '19 at 18:48
  • Perhaps this answer will help. – Blue Jul 23 '19 at 18:49
  • @bounceback The point might be in the domain but sometimes it is still much less interesting than the limit. For example define $f(x)=\frac{sinx}{x}$ when $x\ne 0$ and $f(0)=0$. But the limit when $x\to 0$ is $1\ne f(0)$ in that case. – Mark Jul 23 '19 at 18:50
  • 5
    @bounceback: you substitute the value $x=a$ only when $f$ is continuous at $a$. – Bernard Jul 23 '19 at 18:52
  • Thanks, yes, my apologies – DanLewis3264 Jul 23 '19 at 18:55
  • Well, in calculating the limit we are interested in what $f(x)$ gets close to when $x$ is near $a$ but not equal to $a$. That's a limit. $f(a)$ is a value and may or may not be the same as the limit. But when you say " but after simplifying why we put exactly x=a" I have to confess I have no idea what you mean. We shouldn't do that and so far as I know we do not do that. – fleablood Jul 23 '19 at 18:59
  • 1
    @fleablood I think they mean e.g. simplifying $(x^2-4)/(x-2)$ to $(x+2)$, then plugging in $x=2$ to get a limit of $4$. – Sambo Jul 23 '19 at 19:01
  • Ah, well, that's only one tool you can occasionally use to find a limit if the function has a "removable singularity". If $f(x) = g(x)$ everywhere but $x=a$ but $f(a)$ does not exist but $g(x)$ is continuous at $a$. Then we can prove that as $f(x) = g(x)$ everywhere $x\ne a$. And because $g$ is continuous at $x=a$ so $\lim g(x)=g(a)$ then $\lim_{x\ne a}f(x) = \lim_{x\ne a} g(x) = g(a)$. – fleablood Jul 23 '19 at 22:12

0 Answers0