O que a regra de substituição faz?

Substituting to Evaluate Limits

Indeed, this is a pretty intuitive problem. When x = 5, f(x) = 8. It follows that as x approaches 5 f(x) approaches 8.

Really, what you are doing is a method for evaluating limits: Substitution.

You take, the equation, and substitute the value that x approaches in the limit:

$\lim_{x \rightarrow 5} f(x) = x + 3$

$\lim_{x \rightarrow 5} f(x) = (5) + 3$

$\lim_{x \rightarrow 5} f(x) = 8$

Now onto something a bit trickier:

$\lim_{x \rightarrow -2} (x+2)/(x^2-4)$

Clearly you can see that plugging in two creates a problem:

$\lim_{x \rightarrow -2} (x+2)/(x^2-4) = ((-2)+2)/((-2)^2 -4)$

$\lim_{x \rightarrow -2} (x+2)/(x^2-4) = (4)/(4 -4)$

$\lim_{x \rightarrow -2} (x+2)/(x^2-4) = (4)/(0)$

This limit is actually defined! The function approaches the same value from the left and right as x approaches 2.

Take a look at the graph of the function:

Note the point a, which is f(x) where x = -2. Although the function is undefined when x = -2, the function clearly approaches the same value from both sides. So how do we resolve this paradox?

$\lim_{x \rightarrow -2} (x+2)/(x^2-4) = \lim_{x \rightarrow -2} (x+2)/((x+2)(x-2))$

$\lim_{x \rightarrow -2} (x+2)/(x^2-4) = \lim_{x \rightarrow -2} 1/(x-2)$

By factoring and cancelling, suddenly we are able to substitute!

$\lim_{x \rightarrow -2} (x+2)/(x^2-4) = 1/((-2)-2)$

$\lim_{x \rightarrow -2} (x+2)/(x^2-4) = -1/4$

In this way you see that you sometimes have to do a little extra work for substitution to work.

Remember that it is possible for a limit to be undefined, take this simple example

$\lim_{x \rightarrow 0} 1/x$

Plugging in zero gives us

$\lim_{x \rightarrow 0} 1/x = 1/(0)$

a division by zero! Clearly there is no way to cancel any terms, so the limit is undefined.

Looking at the graph of the function clearly shows why this is undefined:

As 1/x aproaches zero from the left it equals negative infinity, and as it approaches from the right it equals positive infinity. Therefore, the limit is undefined!

Now, give these practice problems a shot!