# Continuity and limits

I am asked to use continuity to evaluate the limit:

$$\lim_{x\to 1} e^{x^2-x}$$

I know that evaluating both the limit and the function will produce the same value, 1, so that tells me that it is continuos at the point $x=1$. This makes sense right? I just wanted to check to see that my reasoning is grounded.

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Since $e^{x^2-x}$ is continuous, the value of the limit is equal to the value of the function at the point. – Yuval Filmus May 18 '12 at 5:10

## 1 Answer

If a function $f(x)$ is continuous at $x=a$, then $$\lim_{x\to a}f(x) = f(a).$$

If $f(x)$ is continuous at $a$, and $g(x)$ is continuous at $f(a)$, then $$\lim_{x\to a}g(f(x)) = g(f(a)).$$

Here you have the function $f(x) = x^2-x$, which is continuous everywhere, and the function $g(x) = e^x$, which is continuous everywhere. So you can compute the limit by evaluation.

Your argument is upside down: you are asked to use continuity to evaluate the limit, but you are trying to conclude that the function is continuous by arguing that the value of the limit equals the value of the function. Not only is that not what you were asked to do, but more importantly: how did you verify that the value of the limit was indeed $1$, without using continuity?

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I am not totally clear on how you separated the original function in to the two functions f(x) and g(x), but I am familiar with the Theorem you stated. – Kurt May 18 '12 at 5:29
@Kurt: You have a composition, because your function first takes $x$ and computes $x^2-x$, and then takes that result and plugs it into the exponential. So we have that our function is of the form $h(x) = e^{f(x)}$, where $f(x)=x^2-x$. That's how I separated them. – Arturo Magidin May 18 '12 at 15:33