Limits and function values

• January 16th 2013, 02:12 PM
Limits and function values
Suppose we wish to compute

$\lim_{x \rightarrow 1} \frac{x^2 + x + 1}{x + 3} \, ,$

where $\small f(x) = \frac{x^2 + x + 1}{x + 3}$. Since limits only care about what happens as we approach $x = 1$, why do we then compute the limit by simply plugging in $x = 1$ in $f(x)$?

What if we were to discover that $f(1) = 1337$ but $f(0.99999)$ going towards negative infinity and $f(1.0000001)$ going towards $\pi$, or something random like that? How can we be so sure of this not happening (without actually checking values close to $x = 1$)?

What's the justification for equating $\lim_{x\rightarrow 1}$ with $f(1)$?
• January 16th 2013, 02:41 PM
emakarov
Re: Limits and function values
Quote:

What's the justification for equating $\lim_{x\rightarrow 1}$ with $f(1)$?

It is that f(x) is continuous at x= 1; therefore, its limit equals its value. The fact that f(x) is continuous is implied by the fact that f(x) is a composition of continuous functions. In particular, it is known that 1 / (x + 3) is continuous everywhere except when the denominator turns into 0.
• January 16th 2013, 02:42 PM
Barioth
Re: Limits and function values
If i remember right, if the fonction is continuous in f(a) then (lim x=> a) = f(a)
• January 16th 2013, 02:49 PM
Re: Limits and function values
Quote:

Originally Posted by emakarov
It is that f(x) is continuous at x= 1; therefore, its limit equals its value. The fact that f(x) is continuous is implied by the fact that f(x) is a composition of continuous functions. In particular, it is known that 1 / (x + 3) is continuous everywhere except when the denominator turns into 0.

Ah, alright I understand! But suppose we didn't know that f(x) was continuous?
• January 16th 2013, 03:13 PM
Barioth
Re: Limits and function values
well you cant not know... you have to check it to evaluate the limit. You probably do it without even knowing.