Checkpoint 1.7.1. Squeeze Theorem Visual.
Graph \(\frac{-9}{x^2+3x}\text{,}\) \(\frac{-5}{x^2+3x}\text{,}\) \(\frac{2\cos(x)-7}{x^2+3x}\) together. See how the third expression remains between the first two as you look to the right.
Section 1.7 Techniques for Evaluating Limits
Standards:
- Evaluate a limit requiring algebraic manipulation
- Evaluate a limit requiring squeeze theorem
When the many mathematical symbols you know began to be used, Thomas Harriot made the statement that mathematics was now covered over in a scab of symbols. As with other problems in mathematics, sometimes the difficulty in evaluating a limit is the form in which we have written it. Specifically, the notation leads to arithmetic with infinity for which sense cannot be made. Modifying that algebraic form by applying algebra, trigonometry, or other ideas enables us to recognize the value of the limit. Our interest here is not so much evaluating the limit (the number or infinity at the end), but rather recognizing how odd infinity is.
We have two goals for this section.
- Learn to use four techniques for handling limits that cannot be evaluated directly using continuity.
- Learn to identify the issue that informs us when to select each technique.
Subsection 1.7.1 Oscillations (and more): Squeeze Theorem
This technique is often used to handle dampened oscillations (think functions with sine or cosine).
We know that a limit does not exist if there are bad oscillations which means the function oscillates infinitely many times in the interval considered and those oscillations do not dampen (amplitude drops to zero). If the oscillations are finite or do dampen, then the limit can exist. One way of showing that a limit with infinite, dampened oscillation exists is to show that it is squeezed (dampened) to a specific value.
Example 1.7.2. How to handle (well-behaved) oscillations.
Evaluate
\begin{equation*}
\lim_{x \to \infty} \frac{2\cos(x)-7}{x^2+3x}\text{.}
\end{equation*}
Note that the cosine will oscillate infinitely. This suggests using the squeeze theorem. We will start with what we know about \(\cos(x)\) and use algebra to convert that into the expression in this limit.
\begin{align*}
-1 \le \amp \cos(x) \le \amp 1\\
-2 \le \amp 2\cos(x) \le \amp 2\\
-9 \le \amp 2\cos(x)-7 \le \amp -5\\
\frac{-9}{x^2+3x} \le \amp \frac{2\cos(x)-7}{x^2+3x} \le \amp \frac{-5}{x^2+3x}\\
\lim_{x \to \infty} \frac{-9}{x^2+3x} \le \amp \lim_{x \to \infty} \frac{2\cos(x)-7}{x^2+3x} \le \amp \lim_{x \to \infty} \frac{-5}{x^2+3x}
\end{align*}
Look at the first step.
- Why did we want a \(cos(x)\text{?}\)
- Note we know the bounds by knowing \(cos(x)\text{:}\) it has nothing to do with this problem.
- The steps were multiply, add, divide in that order. What order is this (what other problems use this order)?
- In the penultimate (second to last) step what is the expression in the middle? Why do we want that for this problem?
For the squeeze theorem to be useful, we need to know the limits on the left and right. We start with the limit on the left for no good reason.
\begin{align*}
\lim_{x \to \infty} x^2+3x = \amp \text{ sum}\\
\lim_{x \to \infty} x^2 + \lim_{x \to \infty} 3x = \amp \text{ scalar}\\
\lim_{x \to \infty} x^2 + 3\lim_{x \to \infty} x = \amp \text{ exponent}\\
\left(\lim_{x \to \infty} x\right)^2 + 3\lim_{x \to \infty} x = \amp \text{ basic forms}\\
= \amp \infty.\\
\lim_{x \to \infty} \frac{-9}{x^2+3x} = \amp \text{ substitution from above}\\
\lim_{u \to \infty} \frac{-9}{u} = \amp \text{ scalar}\\
-9\lim_{u \to \infty} \frac{1}{u} = \amp \text{ basic form}\\
-9 \cdot 0 = \amp 0.
\end{align*}
For the limit on the right, notice that the steps will be the same except for writing -5 instead of -9. The result is still 0. Because our limit must remain below the right limit and above the left limit, and they both go to 0 the squeeze theorem tells us
\begin{equation*}
\lim_{x \to \infty} \frac{2\cos(x)-7}{x^2+3x}=0.
\end{equation*}
This means this was a well-behaved oscillation (infinite oscillations, but damped).
Subsection 1.7.2 Infinity minus infinity: Factoring
This technique is used to evaluate limits involving \(\infty - \infty\) that can be factored. Polynomials are one example, but are not the only time an expression can be factored.
Example 1.7.3. Too much of an infinite thing (Part 1).
In the following limit use of the addition property would leave us with \(\infty-\infty\) which is not informative. Note how factoring removes this difficulty.
\begin{align*}
\lim_{x \to \infty} 9x^2-4x+3 = \amp \text{ notice } \infty-\infty\\
\lim_{x \to \infty} x(9x-4)+3 = \amp \text{ factor}\\
\lim_{x \to \infty} x(9x-4)+3 = \amp \infty \text{ this is now } \infty \cdot \infty
\end{align*}
Subsection 1.7.3 Infinity divided by infinity: Dividing
This technique is used to evaluate limits involving \(\infty/\infty\) or \(0/0\text{.}\) These expressions are problematic for reasons similar to \(\infty - \infty\text{.}\) All the examples at which we will look are rational functions (polynomial divided by a polynomial). Some other examples exist (e.g., substitutions into a rational function like \(x=e^x\)).
This is part of the standard, Evaluate a limit requiring algebraic manipulation. Our goal is to see how algebra removes the difficulty in the initial expression of the function. It is not our goal to memorize shortcuts to solve problems like these. Computers will solve these for us in the future, so memorization is no longer an honored skill.
Example 1.7.4. Too much of an infinite thing (Part 2).
In the following limit use of the quotient property would leave us with \(\infty/\infty\) which is not informative. Note how dividing both expressions removes this difficulty.
\begin{align*}
\lim_{x \to \infty} \frac{x^2+2x+6}{2x^2+2x+1} = \amp \text{ notice } \frac{\infty}{\infty}\\
\lim_{x \to \infty} \frac{1/x^2}{1/x^2} \cdot \frac{x^2+2x+6}{2x^2+2x+1} = \amp \text{ introduce division}\\
\lim_{x \to \infty} \frac{1+\frac{2}{x}+\frac{6}{x^2}}{2+\frac{2}{x}+\frac{1}{x^2}} = \amp \text{ distribute}\\
\lim_{x \to \infty} \frac{1+\frac{2}{x}+\frac{6}{x^2}}{2+\frac{2}{x}+\frac{1}{x^2}} = \amp \frac{1}{2}. \text{ this is now a bunch of zeros}
\end{align*}
Example 1.7.5. Too much of an infinite thing (Part 2b).
Note how dividing both expressions is modified to work with the radical.
\begin{align*}
\lim_{x \to \infty} \frac{x-7}{\sqrt{x^2+x-5}} = \amp \text{ notice } \frac{\infty}{\infty}\\
\lim_{x \to \infty} \frac{1/x}{1/x} \cdot \frac{x-7}{\sqrt{x^2+x-5}} = \amp \text{ introduce division}\\
\lim_{x \to \infty} \frac{\frac{1}{x}(x-7)}{\sqrt{\frac{1}{x^2}(x^2+x-5)}} = \amp \text{ move into radical}\\
\lim_{x \to \infty} \frac{1-\frac{7}{x}}{\sqrt{1+\frac{1}{x}-\frac{5}{x^2}}} = \amp \text{ distribute}\\
\lim_{x \to \infty} \frac{1-\frac{7}{x}}{\sqrt{1+\frac{1}{x}-\frac{5}{x^2}}} = \amp 1. \text{ this is now a bunch of zeros}
\end{align*}
Subsection 1.7.4 Infinity minus infinity: Conjugate
This technique is used to evaluate limits involving \(\infty - \infty\) that include a radical. The radical blocks using the factoring approach. It also blocks the approach that will be learned in Section 2.7.
Example 1.7.6. Too much of an infinite thing (Part 3).
In the following limit use of the addition property (subtraction) would leave us with \(\infty-\infty\) which is not informative. However, we cannot factor this time. Note how the conjugate removes this difficulty.
\begin{align*}
\lim_{x \to \infty} x-\sqrt{x^2+1} = \amp \text{ notice } \infty-\infty\\
\lim_{x \to \infty} \frac{x-\sqrt{x^2+1}}{1} \cdot \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}} = \amp \text{ multiply by conjugate }\\
\lim_{x \to \infty} \frac{x^2-(x^2+1)}{x+\sqrt{x^2+1}} = \amp \text{ distribute and reduce }\\
= \lim_{x \to \infty} \frac{-1}{x+\sqrt{x^2+1}} = \amp \text{ looks like } -1/\infty\\
\lim_{x \to \infty} x+\sqrt{x^2+1} = \amp \infty\\
\lim_{u \to \infty} \frac{-1}{u} = \amp 0. \text{ notice the known form }
\end{align*}
