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Calculus Cameron Geisler

Section 1.1 Calculating Limits

We can evaluate limits using a variety of techniques.
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Subsection 1.1.1 Simple Limits (Direct Substitution)

Example 1.1.1. Direct Substitution with a Linear Function.

Consider \(\lim_{x \to 3} (2x+1)\text{.}\) We know that the function \(f(x)=2x+1\) is a line, and it is never undefined, particularly not at \(x=3\text{,}\) because \(f(3)=2(3)+1=7\text{.}\) Then, we expect that as \(x\) approaches 3, \(2x+1\) gets closer to 7. Then,
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\begin{align*} \lim_{x \to 3} (2x+1) \amp = 2(3)+1\\ \amp = 7 \end{align*}
In other words, we can just plug in \(x=3\) to get the limit value.
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In general, for a linear function, you can just plug in the limit point that you’re approaching.
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Example 1.1.2. Direct Substitution with a Quadratic Function.

Consider \(\lim_{x \to 2} (2x^2+x-1)\text{.}\) We know that this function \(f(x)=2x^2+x-1\) is a parabola. In a similar way, as \(x\) approaches 2, the \(y\)-value will approach the function’s \(y\)-value at \(x=2\text{,}\) which we can get by just plugging in \(x=2\text{,}\)
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\begin{align*} \lim_{x \to 2} (2x^2+x-1) \amp = 2(2)^2 + 2 - 1\\ \amp = 9 \end{align*}
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In general, if you can plug in the limit point and get a number (as opposed to something where you divide by 0), then that will be the limit value. This works almost always. These functions are “nice” in that there are no holes, asymptotes, jumps, or anything strange that would make the limits be more challenging to find.
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Example 1.1.3. Direct Substitution with a Rational Function.

Consider \(\lim_{x \to -1} \frac{x-1}{x^2+2x}\text{.}\) Again, let’s see if we can plug in \(x=-1\text{,}\)
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\begin{gather*} \frac{-1-1}{(-1)^2+2(-1)} = \frac{-2}{-1} = 2 \end{gather*}
This means the limit is just 2,
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\begin{align*} \lim_{x \to -1} \frac{x-1}{x^2+2x} \amp = \frac{-1-1}{(-1)^2+2(-1)}\\ \amp = 2 \end{align*}
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In general, the limit of a polynomial function can be found by just substituting the value that \(x\) is approaching.
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Also, for a rational function, the same rule applies, as long as you don’t divide by 0 when you plug it in.
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Exercise Group 1.1.1. Direct Substitution.

Evaluate each limit.
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(b)
\(\lim_{x \to -1} \frac{x^2 - 5x + 2}{2x^3 + 3x + 1}\)
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Answer.
\(-2\)
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(c)
\(\lim_{x \to 5} \sqrt{\frac{x^2}{x - 1}}\)
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Answer.
\(\frac{5}{2}\)
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(o)
\(\lim_{x\to 0} \frac{3}{\sqrt{16+3x+4}}\)
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Answer.
\(\frac{3\sqrt{5}}{10}\)
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(q)
\(\lim_{x\to 0} \frac{100}{(10x-1)^{11}+2}\)
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Answer.
\(100\)
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Subsection 1.1.2 Limits in Indeterminate Form

Most limits will not be this simple. Instead, when you plug in the limit point, you will get an expression of the form \(\frac{0}{0}\text{,}\) where the denominator is 0 (dividing by 0, which makes it undefined), and the numerator is also 0.
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Subsection 1.1.3 Factoring and Canceling

Example 1.1.6. Factoring to Evaluate a Limit.

Consider the limit,
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\begin{gather*} \lim_{x \to 2} \frac{x^2-5x+6}{x-2} \end{gather*}
The function \(\frac{x^2-5x+6}{x-2}\) is undefined at \(x = 2\text{,}\) because plugging in \(x=2\) leads to dividing by 0. However, it turns out that we can algebraically manipulate it, by factoring the numerator,
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\begin{align*} \frac{x^2 - 5x + 6}{x - 2} \amp = \frac{(x - 3)(x - 2)}{x - 2} \end{align*}
Then, cancelling the common factor of \(x - 2\text{,}\)
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\begin{align*} \amp = x - 3 \end{align*}
Then, since there is no denominator anymore, it turns out we can just plug in \(x=2\) to get the answer. In total,
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\begin{align*} \lim_{x \to 2} \frac{x^2-5x+6}{x-2} \amp = \lim_{x \to 2} \frac{(x - 3)(x - 2)}{x - 2}\\ \amp = \lim_{x \to 2} (x-3)\\ \amp = 2-3 = -1 \end{align*}
This works because the simplified version is equal to the original expression,
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\begin{gather*} \frac{x^2 - 5x + 6}{x - 2} = x - 3 \end{gather*}
They are the same everywhere, except for \(x=2\) (because then the left-hand side gives an error for dividing by 0, but the right-hand side gives \(2 - 3 = -1\)). Therefore, if you want to know what \(\frac{x^2-5x+6}{x-3}\) is doing near \(x=2\text{,}\) you can instead consider the simplified expression \(x-3\text{.}\)
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This is because a limit only cares about values of \(x\) near \(x = 2\text{,}\) and doesn’t care about what happens exactly at \(x=2\text{.}\) Thus, in terms of limits, the expressions are equal.
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Exercise Group 1.1.2. Factoring and Canceling.

Evaluate each limit.
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(a)
\(\lim_{x \to -6} \frac{x^2 - 36}{x + 6}\)
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Answer.
\(-12\)
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(b)
\(\lim_{x \to 6} \frac{x^2 - 4x - 12}{x - 6}\)
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Answer.
\(8\)
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(d)
\(\lim_{x \to 2} \frac{x-2}{x^2+x-6}\)
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Answer.
\(\frac{1}{5}\)
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(f)
\(\lim_{x \to 1} \frac{x^4 - 1}{x^3 - 1}\)
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Hint.
Difference of cubes, difference of squares.
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Answer.
\(\frac{4}{3}\)
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(g)
\(\lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3}\)
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Answer.
\(4\)
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(h)
\(\lim_{x \to 15} \frac{x^2 + 3x - 270}{x^2 - 225}\)
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Answer.
\(\frac{11}{10}\)
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(j)
\(\lim_{x \to -1} \frac{2x^2 + 5x + 3}{x + 1}\)
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Answer.
\(1\)
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(l)
\(\lim_{x\to 5} \frac{x-5}{x^2-25}\)
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Answer.
\(\frac{1}{10}\)
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(m)
\(\lim_{x\to -3} \frac{x+3}{x^2+4x+3}\)
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Answer.
\(-\frac{1}{2}\)
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(p)
\(\lim_{x\to 1} \frac{x^2+x-2}{x^2-1}\)
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Answer.
\(\frac{3}{2}\)
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(q)
\(\lim_{x\to -1} \frac{x^2+3x+2}{x^2-x-2}\)
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Answer.
\(-\frac{1}{3}\)
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(r)
\(\lim_{x\to -2} \frac{-2x-4}{x^3+2x^2}\)
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Answer.
\(-\frac{1}{2}\)
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(s)
\(\lim_{x\to 0} \frac{5x^3+8x^2}{3x^4-16x^2}\)
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Answer.
\(-\frac{1}{2}\)
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(t)
\(\lim_{x\to 2} \frac{x^3-8}{x^4-16}\)
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Answer.
\(\frac{3}{8}\)
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(u)
\(\lim_{x \to -4} \frac{16 - x^2}{x^3 + 64}\)
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Answer.
\(\frac{1}{6}\)
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(v)
\(\lim_{x \to 4} \frac{x^2 - 16}{x^2 - 5x + 4}\)
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Answer.
\(\frac{8}{3}\)
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Exercise Group 1.1.3. More Factoring Practice.

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Subsection 1.1.4 Rationalizing (Removing Roots)

Example 1.1.7. Rationalizing the Numerator.

Consider the limit,
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\begin{gather*} \lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x} \end{gather*}
If we try to plug in \(x=0\text{,}\) we get \(\frac{0}{0}\text{,}\) which is undefined. To simplify this expression, we can rationalize the numerator. To do this, multiply by the conjugate, which is \(\sqrt{4+x}+2\text{,}\) in both the numerator and denominator,
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\begin{gather*} = \lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x} \cdot \frac{\sqrt{4+x} + 2}{\sqrt{4+x} + 2} \end{gather*}
In the numerator, multiplying together the conjugates gives a difference of squares, which cancels out the square root,
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\begin{align*} \amp = \lim_{x \to 0} \frac{(\sqrt{4+x})^2 - 2^2}{x(\sqrt{4+x} + 2)}\\ \amp = \lim_{x \to 0} \frac{4 + x - 4}{x(\sqrt{4+x} + 2)}\\ \amp = \lim_{x \to 0} \frac{x}{x(\sqrt{4+x} + 2)} \amp\amp \text{simplifying the numerator}\\ \amp = \lim_{x \to 0} \frac{1}{\sqrt{4+x} + 2} \amp\amp \text{cancelling the common factor of } x \end{align*}
Now, we can plug in \(x=0\text{,}\)
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\begin{align*} \amp = \frac{1}{\sqrt{4+0} + 2}\\ \amp = \frac{1}{4} \end{align*}
Therefore, the answer is \(\frac{1}{4}\text{.}\)
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In general, rationalize by multiplying by the conjugate, which is the same expression except the middle sign is flipped. This takes advantage of the difference of squares pattern,
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\begin{align*} (a-b)(a+b) \amp = a^2-b^2 \end{align*}
To get rid of the square root.
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Exercise Group 1.1.4. Project I.

Evaluate each limit.
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(a)
\(\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}\)
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Answer.
\(\frac{1}{2}\)
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(b)
\(\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}\)
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Answer.
\(\frac{1}{4}\)
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(c)
\(\lim_{x \to 0} \frac{2}{\sqrt{3x + 4} + 2}\)
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Answer.
\(\frac{1}{2}\)
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(d)
\(\lim_{x \to 0} \frac{2 - \sqrt{4 + x}}{x}\)
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Answer.
\(-\frac{1}{4}\)
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(e)
\(\lim_{x\to 1} \frac{\sqrt{x}-1}{x-1}\)
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Answer.
\(\frac{1}{2}\)
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(f)
\(\lim_{x\to 49} \frac{\sqrt{x}-7}{x-49}\)
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Answer.
\(\frac{1}{14}\)
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(g)
\(\lim_{x\to 0} \frac{\sqrt{x+1}-1}{x}\)
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Answer.
\(\frac{1}{2}\)
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(h)
\(\lim_{x\to 0} \frac{3}{\sqrt{3x+1}+1}\)
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Answer.
\(\frac{3}{2}\)
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(i)
\(\lim_{x\to 0} \frac{\sqrt{5x+4}-2}{x}\)
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Answer.
\(\frac{5}{4}\)
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(j)
\(\lim_{x\to 9} \frac{\sqrt{x}-3}{x-9}\)
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Answer.
\(\frac{1}{6}\)
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(m)
\(\lim_{x\to 3} \frac{2-\sqrt{7-x}}{x-3}\)
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Answer.
\(\frac{1}{4}\)
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Exercise Group 1.1.5. Project II.

Evaluate each limit.
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(a)
\(\lim_{x \to 0} \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x}\)
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Answer.
\(-\frac{\sqrt{7}}{7}\)
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(c)
\(\lim_{x \to 4^+} \frac{x^2 - 16}{\sqrt{x - 4}}\)
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Hint.
Note that the limit is a right-hand limit, because the expression is not defined for \(x \lt 4\text{.}\)
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Answer.
\(0\)
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(d)
\(\lim_{x\to 0} \frac{\sqrt{x^2+9}-3}{x^2}\)
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Answer.
\(\frac{1}{6}\)
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(e)
\(\lim_{x\to 4} \frac{4x-x^2}{2-\sqrt{x}}\)
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Answer.
\(16\)
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(f)
\(\lim_{x\to -1} \frac{\sqrt{x^2+8}-3}{x+1}\)
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Answer.
\(-\frac{1}{3}\)
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(g)
\(\lim_{x\to 2} \frac{\sqrt{x^2+12}-4}{x-2}\)
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Answer.
\(\frac{1}{2}\)
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(h)
\(\lim_{x\to -2} \frac{x+2}{\sqrt{x^2+5}-3}\)
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Answer.
\(-\frac{3}{2}\)
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(i)
\(\lim_{x\to -3} \frac{2-\sqrt{x^2-5}}{x+3}\)
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Answer.
\(\frac{3}{2}\)
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(j)
\(\lim_{x\to 4} \frac{4-x}{5-\sqrt{x^2+9}}\)
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Answer.
\(\frac{5}{4}\)
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Subsection 1.1.5 Limits with Absolute Values

Example 1.1.8. Absolute Value Limit at Zero.

Consider the limit,
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\begin{gather*} \lim_{x \to 0} \frac{|x|}{x} \end{gather*}
Graphically, the function looks like: Graph of \(\frac{|x|}{x}\). Notice that there is a jump at \(x=0\text{.}\) As \(x\) approaches 0 from the right, the function approaches 1, while as \(x\) approaches 0 from the left, the function approaches \(-1\text{.}\) Therefore, the two-sided limit does not exist.
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Algebraically, if you try to plug in \(x=0\text{,}\) you get \(\frac{0}{0}\text{,}\) which is indeterminate. To evaluate this limit, we need to consider the left-hand and right-hand limits separately, because the absolute value function behaves differently on each side of 0.
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Recall that the absolute value function is defined as,
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\begin{align*} |x| = \begin{cases} x \amp x \geq 0 \\ -x \amp x \lt 0 \end{cases} \end{align*}
  • From the right, as \(x \to 0^+\text{,}\) \(|x| = x\) (because \(x\) is positive), so,
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    \begin{align*} \lim_{x \to 0^+} \frac{|x|}{x} \amp = \lim_{x \to 0^+} \frac{x}{x}\\ \amp = \lim_{x \to 0^+} 1 \amp\amp \text{cancelling } x\\ \amp = 1 \end{align*}
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  • From the left, as \(x \to 0^-\text{,}\) \(|x| = -x\) (because \(x\) is negative), so,
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    \begin{align*} \lim_{x \to 0^-} \frac{|x|}{x} \amp = \lim_{x \to 0^-} \frac{-x}{x}\\ \amp = \lim_{x \to 0^-} -1 \amp\amp \text{cancelling } x\\ \amp = -1 \end{align*}
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Since these one-sided limits are not equal, the limit does not exist.
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Example 1.1.9. Absolute Value Limit at a Nonzero Point.

Consider the limit,
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\begin{gather*} \lim_{x \to 2} \frac{|x-2|}{x-2} \end{gather*}
This is similar to the previous example, except the jump is at \(x=2\) instead of \(x=0\text{.}\) Graphically, the function looks like: Graph of \(\frac{|x-2|}{x-2}\).
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Algebraically,
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\begin{align*} \lim_{x \to 2^+} \frac{|x-2|}{x-2} \amp = \lim_{x \to 2^+} \frac{x-2}{x-2}\\ \amp = \lim_{x \to 2^+} 1\\ \amp = 1 \end{align*}
For the left limit,
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\begin{align*} \lim_{x \to 2^-} \frac{|x-2|}{x-2} \amp = \lim_{x \to 2^-} \frac{-(x-2)}{x-2}\\ \amp = \lim_{x \to 2^-} -1 \amp\amp \text{cancelling the common factor of } x-2\\ \amp = -1 \end{align*}
Since these one-sided limits are not equal, the two-sided limit does not exist.
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Exercise Group 1.1.6. Absolute Value Limits.

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Subsection 1.1.6 Simplifying Complex Fractions

These limits often come from the derivative of a function.
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Exercise Group 1.1.7. Complex Fractions.

Find each limit.
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(b)
\(\lim_{x\to 0} \frac{\frac{1}{x-1}+\frac{1}{x+1}}{x}\)
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Answer.
\(-2\)
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(c)
\(\lim_{x\to 0} \frac{\frac{1}{5+x}-\frac{1}{5}}{x}\)
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Answer.
\(-\frac{1}{25}\)
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(d)
\(\lim_{x\to 1} \left(\frac{1}{x^2-x}-\frac{1}{x-1}\right)\)
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Answer.
\(-1\)
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(e)
\(\lim_{x\to 5} \left(\frac{1}{x^2-4x-5}-\frac{1}{6(x-5)}\right)\)
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Answer.
\(-\frac{1}{36}\)
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(f)
\(\lim_{x\to 3} \left(4x-\frac{2}{x-3}\right)(6+x-x^2)\)
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Answer.
\(10\)
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Subsection 1.1.7 One-Sided Limits

Checkpoint 1.1.10. Piecewise Function Limit.

Let \(f(x)=\begin{cases} x^2 \amp x \lt 1 \\ -x+2 \amp x \geq 1 \end{cases}\text{.}\) Find \(\lim_{x \to 1} f(x)\text{,}\) if it exists.
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Answer.
\(1\)
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Checkpoint 1.1.11. Another Piecewise Function Limit.

Let \(f(x)=\begin{cases} 8-2x \amp x \lt 4 \\ \sqrt{x-4} \amp x \gt 4 \end{cases}\text{.}\) Find \(\lim_{x \to 4} f(x)\text{,}\) if it exists.
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Answer.
\(0\)
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Subsection 1.1.8 Substitution (Change of Variable)

Exercise Group 1.1.8. Change of Variable.

Evaluate the limit by using a change of variable.
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(a)
\(\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8}\)
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Answer.
\(\frac{1}{12}\)
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(b)
\(\lim_{x \to 0} \frac{(x+8)^{1/3} - 2}{x}\)
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Answer.
\(\frac{1}{12}\)
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(c)
\(\lim_{x \to 27} \frac{27 - x}{x^{1/3} - 3}\)
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Answer.
\(-27\)
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(d)
\(\lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1}\)
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Answer.
\(6\)
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(e)
\(\lim_{x \to 1} \frac{x^{1/6} - 1}{x^{1/3} - 1}\)
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Answer.
\(2\)
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(f)
\(\lim_{x \to 4} \frac{\sqrt{x} - 2}{\sqrt{x^3} - 8}\)
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Answer.
\(\frac{1}{12}\)
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