Calcluating Limits at Finite Values

Section 4.2 Calcluating Limits at Finite Values

Subsection 4.2.1 Three Steps

There is a three step procedure to calculating limits.

First, try to evaluate the function at the limit point. For our standard elementary functions (polynomials, roots, trig, exponentials, logarithms, etc.), the limit at a point in the domain is just the value of the function. This is a property called continuity, which I will discuss in more detail in Section 4.3.
If the first method fails, try to analyze the piece of the limit to see if a direct conclusion is possible. Even though all three steps have logic to them, I’ll call this the logical analysis step.
If both methods fail, the limit is called an indeterminate form. In this case, I need to use various algebraic tricks (factoring, expanding, multiplying by conjugates, trig identities) and rules for manipulating limits to change the limit into a more approachable form.

Subsection 4.2.2 Limit Rules

The third step referred to the rules for manipulating limits, which I will now state. These rules mostly tell me that limits work well with the standard operations of arithmetic. Let \(f\) and \(g\) be functions with limits defined at \(x=a\) and let \(c\) be a constant. (In the quotient limit, assume \(g(x) \neq 0\text{,}\) and in the exponential limit, assume \(f(x) > 0\text{.}\))

\begin{align*} \lim_{x \rightarrow a} f(x) + g(x) \amp = \lim_{x \rightarrow a} f(x) + \lim_{x \rightarrow a} g(x)\\ \lim_{x \rightarrow a} f(x) - g(x) \amp = \lim_{x \rightarrow a} f(x) - \lim_{x \rightarrow a} g(x)\\ \lim_{x \rightarrow a} c f(x) \amp = c \lim_{x \rightarrow a} f(x)\\ \lim_{x \rightarrow a} f(x) g(x) \amp = \lim_{x \rightarrow a} f(x) \lim_{x \rightarrow a} g(x)\\ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \amp = \frac{\lim_{x \rightarrow a} f(x)} {\lim_{x \rightarrow a} g(x)}\\ \lim_{x \rightarrow a} (f(x))^c \amp = (\lim_{x \rightarrow a} f(x))^c \end{align*}

In addition to the limit rules, there are two special limits that are frequently useful. The second limit can be taken as a definition of the exponential base \(e\text{.}\)

\begin{align*} \lim_{x \rightarrow 0} \frac{\sin x}{x} \amp = 1\\ \lim_{x \rightarrow 0^+} \left( 1 + x \right)^{\frac{1}{x}} \amp = e \end{align*}

Subsection 4.2.3 Examples

In the examples, I demonstrate each of the three steps in the limit calculation process, as well as the use of the limit rules to splits limits up over arithmetic.

Example 4.2.1.

The limit

\begin{equation*} \lim_{x \rightarrow 3} \frac{x^2 + 9x - 15}{x+7} \end{equation*}

can be directly evaluated. The function is perfectly well defined at \(x=3\text{.}\) The value of the limit is just the value of the function.

\begin{equation*} \lim_{x \rightarrow 3} \frac{x^2 + 9x - 15}{x+7} = \frac{3^2 + 9(3) - 15}{3 + 7} = \frac{21}{10} \end{equation*}

Example 4.2.2.

The limit

\begin{equation*} \lim_{x \rightarrow -2} \frac{x+4}{x+2} \end{equation*}

cannot be directly evaluated due to division by zero. However, I can work this out logically. The numerator approaches \(-2 + 4 = 2\text{,}\) so I can think of it as some number very close to \(2\text{.}\) The denominator is becoming very close to zero. If I take a number very close to \(2\) and divide by a very small number, I get a very large number. As I divide by smaller and smaller numbers, I get larger and larger numbers. I conclude that the limit diverges, getting larger and larger without bound. Since \(x \rightarrow -2\) comes from both sides of the numberline, the denominator (and hence the whole limit) could be either positive and negative, depending on the side of approach. I can write this conclusion in a couple ways. First, I can explicitly show that this diverges to either positive or negative infinity (that this grows without bound either in the positive or negative direction depending on which side of the limit is considered).

\begin{equation*} \lim_{x \rightarrow -2} \frac{x+4}{x+2} = \pm \infty \end{equation*}

Second, we could also simply write that the limit doesn’t exist. This is less precise, but still correct.

\begin{equation*} \lim_{x \rightarrow -2} \frac{x+4}{x+2} \text{ DNE} \end{equation*}

Example 4.2.3.

The limit

\begin{equation*} \lim_{x \rightarrow 7} \frac{x^2 - 4x - 21}{x^2 + 3x - 70} \end{equation*}

cannot be evaluated due to division by zero. I also can’t work this out by just logical analysis of the pieces, since both the numerator and denominator are approaching zero. This is an indeterminate form, so I need to do some algebra to figure it out. I’m going to factor both quadratics.

\begin{equation*} = \lim_{x \rightarrow 7} \frac{(x-7)(x+3)}{(x-7)(x+10)} \end{equation*}

Then I can cancel of the \((x-7)\) terms. This is going to remove the division by zero problem, which is my goal. With indeterminate forms, I’m always trying to remove the problem that prevents me from either evaluating directly or using logic analysis of the pieces to determine the limit.

\begin{equation*} = \lim_{x \rightarrow 7} \frac{x+3}{x+10} \end{equation*}

Now I can just evaluate the limit, since there is no more division by zero.

\begin{equation*} = \frac{10}{17} \end{equation*}

Example 4.2.4.

\begin{equation*} \lim_{x \rightarrow 1} x \left( \frac{x^2-1}{x-1} - \frac{3}{x-4} \right) \end{equation*}

This limit has various pieces combined together with arithmetic. I can use the limit rules to split it up. First, I’ll split up the product.

\begin{equation*} = \left( \lim_{x \rightarrow 1} x\right) \lim_{x \rightarrow 1} \left( \frac{x^2-1}{x-1} - \frac{3}{x-4} \right) \end{equation*}

Then the first limit evaluates to \(1\text{.}\) I’ll use the limit rules again to split up the difference in the second limit.

\begin{equation*} = 1 \left( \lim_{x \rightarrow 1} \frac{x^2-1}{x-1} - \lim_{x \rightarrow 1} \frac{3}{x-4} \right) \end{equation*}

The second limit can be directly evaluated. The first limit needs some work, so I factor its numerator.

\begin{equation*} = \left( \lim_{x \rightarrow 1} \frac{(x-1)(x+1)}{x-1} - \frac{3}{-3} \right) \end{equation*}

Then I can cancel the \((x-1)\) term in the remaining limit.

\begin{equation*} = \lim_{x \rightarrow 1} (x + 1) + 1 \end{equation*}

Finally, I can now evaluate the remaining limit and finish the whole problem.

\begin{equation*} = 2 + 1 = 3 \end{equation*}

Example 4.2.5.

The limit

\begin{equation*} \lim_{x \rightarrow 0} \frac{x^2 + \sin x}{x} \end{equation*}

is an indeterminate form. I can use the limit rules to break it into two limits.

\begin{equation*} = \lim_{x \rightarrow 0} \frac{x^2}{x} + \lim_{x \rightarrow 0} \frac{\sin x}{x} \end{equation*}

In the first limit, I can cancel an \(x\text{.}\) The second is one of the two special limits I mentioned above, so I just use its known value. Then I can evaluate the last remiaining limit to finish the problem.

\begin{equation*} = \lim_{x \rightarrow 0} x + 1 = 0 + 1 = 1 \end{equation*}

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