Limit Problems of Polynomials

When you're talking about the limit of a ratio of two polynomials, it's really important to notice where the limit is being taken. In particular, the techniques for dealing with \(x \to \infty\) and \(x \to a\) (some finite \(a\)) are completely different.

Limits Involving Rational Functions as \(x \to a\)

If the denominator is not \(0\), you can just substitute the variable \(x\) with its value:

\[ \lim_{x \to -1} \frac{x^2 - 3x + 2} {x - 2} = \frac{(-1)^2 - 3(-1) + 2}{-1 - 2} = \frac{6}{-3} = -2 \]

But if you want to find:

\[ \lim_{x \to 2} \frac{x^2 - 3x + 2} {x - 2} \]

the substitutiong method may not work, you'll get \((4 - 6 + 2) / (2 - 2)\), which is \(0 / 0\). This is called an indeterminate form. The technique we used to solve this is called factoring:

\[ \lim_{x \to 2} \frac{x^2 - 3x + 2} {x - 2} = \lim_{x \to -2} \frac{(x - 2)(x - 1)} {x - 2} = \lim_{x - 2}(x - 1) \]

Now we plug \(x = 2\) into the expression \((x - 1)\), then we get \(1\). That's the value of the limit we're looking for.

In addition to knowing how to factor quadratics, it's really useful to know the formula for the difference of two cubes:

\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]

But what if the denominator is \(0\) but the numerator is not \(0\)? There are four types of behavior that could arise:

Limits Involving Square Roots as \(x \to a\)

Consider the following limit:

\[ \lim_{x \to 5} \frac{\sqrt{x^2 - 9} - 4} {x - 5} \]

If you plug in \(x = 5\), you get the indeterminate form \(0 / 0\). What you need to do is multiply and divide by \(\sqrt{x^2 - 9} + 4\), which is called conjugate expression of \(\sqrt{x^2 - 9} - 4\):

\[ \lim_{x \to 5} \frac{\sqrt{x^2 - 9} - 4} {x - 5} = \lim_{x \to 5} \frac{\sqrt{x^2 - 9} - 4} {x - 5} \times \frac{\sqrt{x^2 - 9} + 4} {\sqrt{x^2 - 9} + 4} = \lim_{x \to 5} \frac{x + 5} {\sqrt{x^2 - 9} + 4} = \frac{10}{8} \]

Limits Involving Rational Functions as \(x \to \infty\)

In symbols, we are now trying to find limits of the form:

\[ \lim_{x \to \infty} \frac{p(x)}{q(x)} \]

where p and q are polynomials.

The very important property of a polynomial:

When x is large, the leading term dominates.

And we have the theoram:

\[ \lim_{x \to \infty} \frac{C}{x^n} = 0 \]

For example, the limit is solved:

\[ \lim_{x \to \infty} \frac{3x^3 - 1000x^2 + 5x - 7} {3x^3} = \lim_{x \to \infty}(1 - \frac{1000}{3x} + \frac{5}{3x^2} - \frac{7}{3x^3}) = 1 - 0 + 0 + 0 = 1 \]

Limits Involving Poly-type Function as \(x \to \infty\)

The principles for poly-type functions are similar to those for polynomials, except that this time it may not be so clear what the leading term is.

Limits Involving Rational Functions as \(x \to -\infty\)

Now we solve the limits of the form:

\[ \lim_{x \to \infty} \frac{p(x)} {q(x)} \]

where p and q are polynomials or even poly-type functions.

All the principles we've been using apply equally well here.

The only situation we need to pay attention is:

if \(x < 0\) and you want to write \(^n\sqrt{x^y} = x^m\), the only time you need a minus sign in front of \(x^m\) is when \(n\) is even and m is odd.