Section 2.2 · Limits

Overview

We can compute limits using graphs, using tables, and using algebra. Regardless of method, the idea of a limit is to examine how the function behaves as it approaches a particular \(x\)-value. A two-sided limit exist only when the function approaches the same value from the left and right.

Computing limits from graphs

In general, we can compute \(\lim_{x\to a} f(x)\) from a graph of \(f(x)\) by using the graph to understand how the function behaves on both sides of \(x=a\).

Left probe: x→a⁻, y ≈ Right probe: x→a⁺, y ≈ f(a): Conclusion:

Drag the blue dots toward \(a\) (the \(x\)-coordinate of the red dot) from left and right. Shrink the shaded window around \(a\). If the left and right y-values get close to the same number, that’s the limit. If they head to different numbers (or blow up or keep oscillating), the limit does not exist. The value at \(a\) (if any) doesn’t have to equal the limit.

Example Let \(f(x)\) be the function graphed below.
Compute \( \displaystyle \lim_{x\to -2} f(x)\), if it exists. Then find \(f(-2)\).
Solution
  • Left-hand limit: \(\displaystyle\lim_{x\to -2^-} f(x)=1.5\) since the curve approaches the open circle’s height.
  • Right-hand limit: \(\displaystyle\lim_{x\to -2^+} f(x)=1.5\).
  • Therefore \(\displaystyle\lim_{x\to -2} f(x)=1.5\) since \(\displaystyle\lim_{x\to -2^-}f(x)=\lim_{x\to -2^+}f(x)=1.5\).
  • Function value: the filled dot gives \(f(-2)=2.5\) the \(y\)-value of the solid point.

Computing limits from tables

Use \(x\) values approaching \(a\) from both sides. If the value of \(f(x)\) settles, that’s the limit.

Example: Trig function Use a table to compute \( \lim_{x\to 0} f(x)=\dfrac{\cos(2x)-\cos x}{x^2} \).
Solution
  1. Pick inputs: \(x=\{-0.1,-0.01,-0.001,0.001,0.01,0.1\}\).
  2. Compute \(f(x)\) for each value.
  3. Observe the trend. The values approach \(-1.5\), so the limit is \(-\tfrac{3}{2}\).
xf(x)
-0.1-1.493758743678
-0.01-1.499937500875
-0.001-1.499999375043
0.001-1.499999375043
0.01-1.499937500875
0.1-1.493758743678

Computing limits algebraically

Example: Piecewise function Consider the function:

\( f(x)=\begin{cases} x^2-1,& x<2\\ 3x-5,& x\ge 2 \end{cases} \)

Compute \(\lim_{x \to 2} f(x)\), if it exists.
Solution

At \(x=2\):

  • \(\lim_{x\to 2^-} f(x)=(2)^2-1=3\)
  • \(\lim_{x\to 2^+} f(x)=3(2)-5=1\)
Since \(3\neq1\), \(\lim_{x\to2}f(x)\) DNE.

Example: Absolute Value Consider the function:

\( f(x)=\dfrac{|x|}{x} \),

Compute \(\lim_{x\to 0} f(x)\), if it exists.

Solution

At \(x=0\):

  • \(\lim_{x\to 0^-} \frac{|x|}{x} = \lim_{x\to 0^-} \frac{-x}{x} = -1\)
  • \(\lim_{x\to 0^+} \frac{|x|}{x} = \lim_{x\to 0^+} \frac{x}{x} = 1 \)
Since \(-1 \neq1\), \(\lim_{x\to 0}f(x)\) DNE.