Suppose I'm trying to show that $$\lim_{x \rightarrow 2}x^2=4.$$ Obviously this limit exits, but suppose I didn't know that it did. Is there a way I can show that the limit exists without using an epsilon-delta proof or a graph? Would say $\lim_{x \rightarrow 2}x^2=(2)^2=4$ be enough to say that the limit exists or could I say that since $x^2$ is continuous on $(-\infty,\infty)$ then the limit as $x$ approaches $2$ eixsts?

  • $\begingroup$ If you already have proved that polynomials are continuous obviously you know not only that the limit exists, but also that its value is $2^2=4$. $\endgroup$ – Raffaele Oct 20 '17 at 14:50
  • $\begingroup$ you can use that $$x^2-4=(x-2)(x+2)$$ and that $$\lim_{x\to 2}(x-2)(x+2)=0$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 20 '17 at 14:52
  • $\begingroup$ The $\epsilon-\delta$ proof method is a rigorous definition for limit, so I don't think you can really avoid it for a rigorous proof. Any method you use would have it hidden subtly within it. $\endgroup$ – Prasun Biswas Oct 20 '17 at 14:53
  • $\begingroup$ @PrasunBiswas You can avoid it by using other theorems that use it (thus, you implicitly use $\epsilon-\delta$, but you avoid using it as a proof method. $\endgroup$ – 5xum Oct 20 '17 at 14:59
  • $\begingroup$ The $\epsilon, \delta$ definition of limits is not a practical method to evaluate limits. Limits are almost always evaluated using theorems meant to evaluate limits (algebra of limits, squeeze theorem, substitution, L'Hospital's Rule, Taylor expansions etc) combined with a certain set of standard limits. The limit in question is very easy and uses two key ideas: the standard limit $\lim_{x\to a} x=a$ (proved easily via definition) and the product rule of limits. $\endgroup$ – Paramanand Singh Oct 20 '17 at 15:55

If you already know:

  1. That $x\mapsto x^2$ is continuous on $(-\infty, \infty)$
  2. The theorem:

    $f$ is continuous at $a$ if and only if $$\lim_{x\to a} f(x)=f(a)$$

then sure, you can say $$\lim_{x\to 2} x^2=2^2.$$

But if you don't yet know points $1$ and $2$, then you cannot.


using epsilon-delta proof method is the way you define limit so for not using it you will have to use a method that use it in the background. In that case, a function is continuous at $a$ iff $\lim_{x\to a}f(x)=f(a)$, and you know that $x^2$ is continuous at $(-\infty,\infty)$ so you can say that for $f(x)=x^2$ you have $\lim_{x\to 2}f(x)=2^2=4$, but if you want to show that $x^2$ is continuous over $(-\infty,\infty)$ you will have to use the epsilon-delta proof


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