Calculus Blog: September 2012

Friday, September 14, 2012

Limits Involving Infinity

It's hard to follow up the Sandwich Thm...but I'll try. :) HW help video at bottom!

Limits involving infinity really revolve around finding horizontal asymptotes and vertical asymptotes.

If you are trying to find a limit as "x" goes to infinity (or negative infinity), you are lookign for horizontal aysmpotes. The key? Look at the powers:
1. Degree on bottom is bigger: Limit = 0

Because the denominator's largest power is larger than the numerator's largest power, the denominator's highest powered term takes this expression over as x approaches inifinity. The denominator becomes larger much quicker than the denominator, therefore the limit of this expression must be zero. Because zero is a constant, this means that there is a horizontal asymptote at

2. Degree on top = degree on bottom: Limit = A/B (where A = leading coefficient on top, B = leading coefficient on bottom)

In this limit problem, the highest power term in the numerator is

and the highest power term in the denominator is

. Since these are the highest power terms, they dominate the limit problem, and we can ignore the other terms in determining the limit. We see that the limit is the quotient of the coefficients of the highest power terms.

3. Degree on top is bigger: Divide everything by biggest power on the bottom, and let x go to infinity of negative infinity and see where your function goes.

Because the numerator's largest power is larger than the denominator's largest power, the numerator's highest powered term takes this expression over as x approaches infinity. Therefore the limit of this type of expression must be positive infinity.

The Sandwich (or Squeeze) Theorem

The Sandwich Thm is a very creative way to find limits of functions that are down right impossible to solve using algebriac techniques. Many times, sandwich thm problems involve trig functions...although they don't have to. This is because we can easily "start" the problem be sandwiching sine or cosine between 1 and -1...as you no doubt recall from trig. :)

As a motivation let us consider the function

$\begin{displaymath}f(x) = x^2 \sin\left(\frac{1}{x}\right)\;\cdot\end{displaymath}$

When x get closer to 0, the function $\displaystyle g(x)=\sin\left(\frac{1}{x}\right)$ fails to have a limit. So we are not able to use the basic properties discussed in the previous pages. But we know that this function $\displaystyle g(x)=\sin\left(\frac{1}{x}\right)$ is bounded below by -1 and above by 1, i.e.

$\begin{displaymath}-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\end{displaymath}$

for any real number x. Since $x^2 \geq 0$ , we get

$\begin{displaymath}- x^2 \leq x^2\;\sin\left(\frac{1}{x}\right) \leq x^2 \;. \end{displaymath}$

Hence when x get closer to 0, x² and -x² become very small in magnitude. Therefore any number in between will also be very small in magnitude. In other words, we have

$\begin{displaymath}\lim_{x \rightarrow 0}\quad x^2\;\sin\left(\frac{1}{x}\right) = 0\;.\end{displaymath}$

Enjoy!