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Lecture 3: Finding Limits Analytically

GOAL: Compute limits without guessing.

We start with non-piecewise functions.

To find \( \lim_{x \to a} f(x) \):

Case 1:

\( f(a) \) is well-defined [i.e., not undefined, not \(\infty\), not \(-\infty\)]. Here, the limit is just \(f(a)\).

Example 1

\[ \lim_{x \to 1} (x^2 + x - 2) = 1^2 + 1 - 2 = 0 \]
Red parabola with a blue tangent-like line through the marked point on the curve at x equals 1.
Figure 1
A hand-drawn red sketch of the parabola \( y = x^2 + x - 2 \) on a set of axes. The curve opens upward, crossing the horizontal axis at approximately \( x = -2 \) (labeled -2 on the left) and at \( x = 1 \). A blue steeply rising line passes through the point on the curve at \( x = 1 \), where a solid blue dot marks the point \( (1, 0) \), illustrating that the function value equals the limit there.

In this example, we say \( f(x) \) is continuous at \( x = 1 \) (we will talk more about continuous functions next lecture). In general,

\[ \lim_{x \to a} f(x) = f\left( \lim_{x \to a} x \right) = f(a) \]

Case 2:

\( f(a) \) takes the form \( \frac{[\text{Non-Zero}]}{0} \). Then the limit is either \( \infty \), \( -\infty \), or DNE. To determine the limit, we look at the sign of the function around \(a\).

Example 2

To compute \(\lim_{x \to 0} \frac{1}{2x}\), we break it up into its left and right limits. Then we look at whether \(f(x)\) is positive or negative to the left and right of \(0\) [Make sure to be close to 0]. The results are in the sign chart below

Sign chart for \( \frac{1}{2x} \)
\( x < 0 \) 0 \( x > 0 \)
Top + +
Bottom − +
Fraction \(\frac{1}{2x}\) − +
Here, \(\frac{1}{2x}\) is negative to the left of \(0\) and positive to the right of \(0\). So the graph looks similar to what is in Figure 2 (near 0).
Sketch of one over two x with a branch rising upward left of the vertical asymptote at x equals 0 and falling below.
Figure 2
A hand-drawn red sketch on horizontal and vertical axes showing the behavior of \( \frac{1}{2x} \) near \( x = 0 \). A thick arrow points upward along the left side of the vertical axis and another thick arrow points downward below the horizontal axis just left of the vertical axis, indicating the two-sided infinite behavior. The vertical line is labeled \( x = 0 \) beneath the sketch.
Therefore, \[ \begin{gathered} \lim_{x \to 0^-} \frac{1}{2x} = -\infty \\[0.5em] \lim_{x \to 0^+} \frac{1}{2x} = \infty\\[0.5em] \implies \lim_{x \to 0} \frac{1}{2x}\text{ DNE} \end{gathered} \]

Example 3

We compute \(\lim_{x \to 0} \frac{1}{x^2}\) in a similar way.
Sign chart for \( \frac{1}{x^2} \)
\( x < 0 \) 0 \( x > 0 \)
Top + +
Bottom + +
Sketch of one over x squared with both branches rising to a vertical asymptote at x equals 0.
Figure 4
A hand-drawn sketch in red of the graph of \( \frac{1}{x^2} \) with a horizontal axis and a vertical axis. The curve has two branches, one on each side of the vertical axis, both lying above the horizontal axis. Each branch rises steeply toward a vertical asymptote at \( x = 0 \) and flattens out toward the horizontal axis as \( x \) moves away from zero in either direction, illustrating that the limit as \( x \) approaches 0 is \( \infty \).
Therefore, \[ \begin{gathered}\lim_{x \to 0^-} \frac{1}{x^2} = \infty \\[0.5em] \lim_{x \to 0^+} \frac{1}{x^2} = \infty \end{gathered}\\[0.5em] \implies \lim_{x \to 0} \frac{1}{x^2} = \infty \]

Example 4

We compute \(\lim_{x \to -1} \frac{-5}{(x+1)^2} \) in a similar way.
Sign chart for \(\frac{-5}{(x+1)^2}\)
\( x < -1 \) \( -1 \) \( x > -1 \)
Top − −
Bottom + +
Sketch with a dashed vertical asymptote at x equals negative 1 and both branches heading downward.
Figure 3
A hand-drawn red sketch of \( \frac{-5}{(x+1)^2} \) with a dashed vertical line marking the asymptote at \( x = -1 \), labeled -1 below the horizontal axis. Curved branches on both sides of the dashed line plunge downward toward negative infinity, and a solid vertical line to the right represents the vertical axis.
\[ \begin{gathered} \lim_{x \to -1^-} \frac{-5}{(x+1)^2} = -\infty \\[0.5em] \lim_{x \to -1^+} \frac{-5}{(x+1)^2} = -\infty \\[0.5em]\implies \lim_{x \to -1} \frac{-5}{(x+1)^2} = -\infty \end{gathered} \]

Case 3

\( f(a) \) gets us an indeterminate form \( \left(\text{for example,} \frac{0}{0} \right) \)

Example 5

Compute \(\lim_{x \to 1} \frac{x^3 - 1}{x - 1}\)

Here we need the Difference of Cubes Formula:
\( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \)

\begin{align*} \lim_{x \to 1} \frac{x^3 - 1}{x - 1} =&~ \lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{(x-1)}\\ =&~ \lim_{x \to 1} (x^2+x+1)\\ =&~ 1+1+1\\ =&~ 3 \end{align*}

This example highlights the main strategy we use to deal with limits in this case: Use Algebra to reduce the limit to Case 1 or Case 2.

Example 6

Compute \(\lim_{x \to 5} \frac{x^3 - 5x^2}{(x-5)^2} \).

\[ \lim_{x \to 5} \frac{x^3 - 5x^2}{(x-5)^2} = \lim_{x \to 5} \frac{x^2(x-5)}{(x-5)^2} = \lim_{x \to 5} \frac{x^2}{x-5}\] Here, factoring shows this limit is in Case 2. Using the sign chart below, we see the limit DNE.
Sign chart for \( \frac{x^2}{x-5} \)
\( x < 5 \) 5 \( x > 5 \)
Top + +
Bottom − +

Example 7

Compute \(\lim_{x \to 2} \frac{x^2 - 3x + 2}{x^2 + 2x - 8}\).

\[ \lim_{x \to 2} \frac{x^2 - 3x + 2}{x^2 + 2x - 8} = \lim_{x \to 2} \frac{(x-1)(x-2)}{(x+4)(x-2)} = \lim_{x \to 2} \frac{x-1}{x+4} = \frac{2-1}{2+4} = \frac{1}{6} \]

Piecewise Functions

This is best seen through an example. Suppose

\[ f(x) = \begin{cases} \sin x & \text{if } -\pi \le x \le \pi \\ x - \pi & \text{if } \pi < x \le 2\pi \\ 5 & \text{if } 2\pi < x \le 3\pi \end{cases} \]
Graph of piecewise function: sine wave from negative pi to pi, rising line to 2 pi, constant 5 to 3 pi.
Figure 5
A hand-drawn red graph on x and y axes. The x-axis is labeled with tick marks at \( -\pi \), \( \pi \), \( 2\pi \), and \( 3\pi \); the y-axis is labeled with values \( -1 \), \( 1 \), \( \pi \), and \( 5 \). From \( x = -\pi \) to \( x = \pi \) the curve is one full sine wave, dipping to \( -1 \) between \( -\pi \) and \( 0 \) and rising to \( 1 \) between \( 0 \) and \( \pi \), returning to the x-axis at \( \pi \). From \( x = \pi \) to \( x = 2\pi \) a straight line rises from the x-axis at \( \pi \) up to a filled dot at height \( \pi \) when \( x = 2\pi \). From \( x = 2\pi \) to \( x = 3\pi \) a horizontal segment sits at height 5, beginning with an open circle at \( x = 2\pi \) and ending with a filled dot at \( x = 3\pi \), showing the jump discontinuity at \( x = 2\pi \).

Case 1

Our point of interest is not a “boundary point.” By “boundary point” I mean where the formula changes. So, in the example the boundary points are \( \pi \) and \( 2\pi \).

Example 7

Compute \(\lim_{x \to \frac{\pi}{2}} f(x) \).

\[ \lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \sin x = \sin \frac{\pi}{2} = 1 \]

Case 2

The point is a boundary point.

Example 8

Compute \(\lim_{x \to 2\pi} f(x) \). Since the formula changes, we are forced to break it up into the left and right limits.

\[ \begin{gathered} \lim_{x \to 2\pi} f(x) \begin{cases} \lim_{x \to 2\pi^-} f(x) = \lim_{x \to 2\pi^-} (x - \pi) = 2\pi - \pi = \pi \\[2ex] \lim_{x \to 2\pi^+} f(x) = \lim_{x \to 2\pi^+} 5 = 5 \end{cases} \\[0.5em] \implies \lim_{x \to 2\pi} f(x) \text{ DNE} \end{gathered} \]

Limit Laws

Loosely speaking, if \( \lim_{x \to a} f(x) \) and \( \lim_{x \to a} g(x) \) exist, then the limits “play well” with arithmetic.

In particular,

Example 9

If \( \lim_{x \to 3} f(x) = 5 \) and \( \lim_{x \to 3} g(x) = 2 \), then

\begin{align*} \lim_{x \to 3} \left[ f(x) + g(x) + 3 \right]^2 =&~ \left[ \lim_{x \to 3} \left( f(x) + g(x) + 3 \right) \right]^2\\ =&~ \left[ \lim_{x \to 3} f(x) + \lim_{x \to 3} g(x) + \lim_{x \to 3} 3 \right]^2\\ =&~ \left[ 5 + 2 + 3 \right]^2\\ =&~ 10^2\\ =&~ 100 \end{align*}

Remark The limits need to exist to apply the Limit Laws.

Example 10

Compute \(\lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{x + x^2} \right)\)

Notably, since the limits of each term DNE.

\[ \lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{x + x^2}\right) \neq \lim_{x \to 0} \frac{1}{x} - \lim_{x \to 0} \frac{1}{x + x^2} \]

We need to do some algebra before we can compute the limit (mainly combining the fractions together).

\begin{align*} \lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{x + x^2} \right) =&~ \lim_{x \to 0} \frac{x + x^2 - x}{x(x + x^2)}\\ =&~ \lim_{x \to 0} \frac{x^2}{x(x + x^2)}\\ =&~ \lim_{x \to 0}\frac{x^2}{x^2(x+1)}\\ =&~ \lim_{x \to 0} \frac{1}{x+1}\\ =&~ 1 \end{align*}