Examples: Computing Limits Algebraically

Example: Constant Function

If \(f(x) = 2\), find \(\lim\limits_{x\to 1^{-}}f(x)\).

Solution

\(f\) is a constant function, hence a polynomial. Therefore, by Property 7 of Properties of Finite Limits, \(\lim\limits_{x\to 1} f(x) = f(1) = 2\). Since the limit exists, the one-sided limits exist, too and coincide with the limit. Thus, \(\lim\limits_{x\to 1^{-}}f(x) = 2\).

Example: Rational Function (1)

Evaluate

\[\lim\limits_{x\to -1} \dfrac{x^3 +4x^2-x-3}{2x^2+1}.\]

Solution

The function

\[ g(x) = \dfrac{x^3 +4x^2-x-3}{2x^2+1}\]

is a rational function, and since \(2(-1)^2 + 1 = 3 \neq 0\), it follows that \(-1\) is in the domain of \(g\). Again by Property 7 of Properties of Finite Limits, \(\lim\limits_{x\to -1} g(x)\) exists and equals

\[g(-1) = \frac{(-1)^3 + 4(-1)^2 - (-1) -3}{2(-1)^2+1} = \frac{1}{3}.\]

Example: Rational Function (2)

Evaluate

\[\lim\limits_{x\to 2} \dfrac{x^2 +4x-1}{x-2}.\]

Solution

For \(x=2\), the denominator of the expression above is zero, so we cannot apply Property 7 of Properties of Finite Limits in this case. We observe that, as \(x\) gets closer to \(2\), the values of the denominator get closer to \(0\), while the values of the numerator approach \(11\).

Moreover, if we approach \(2\) from the left, the values of \((x^2 +4x-1)/(x-2)\) will be negative and approach \(-\infty\), while if we approach from the right, the values are positive and approach \(+\infty\).

Therefore, neither the left-hand limit nor the right-limit exists, and hence \(\lim\limits_{x\to 2} \dfrac{x^2 +4x-1}{x-2}\) does not exist either.

Example: Piecewise Polynomial Function

Evaluate \(\displaystyle\lim_{x\to 6} h(x)\), where

\[\begin{equation*} h(x)= \begin{cases} x^2-4x-12 & \hbox{if } x<6 \\ 8 & \hbox{if } x=6 \\ x^2-5x-6 & \hbox{if } x>6 \\ \end{cases} \end{equation*}\]

Step 1:   Find the limit from the left.

In this case, since we are approaching 6 from the left, we can assume that \(x<6\), and therefore \(h(x) = x^2-4x-12\).

\[\begin{align*} \lim_{x\to 6^{-}} h(x) &= \lim_{x\to 6^-} x^2-4x-12\\ \\ &= 6^2-4(6)-12 && \text{plug in $x=6$ (Property 7)}\\ \\ &= 36-24-12 && \text{simplify}\\ \\ &= 0 \end{align*}\]

Step 2:   Find the limit from the right.

In this case, since we are approaching 6 from the right, we can assume that \(x>6\), and therefore \(h(x) = x^2-5x-6\).

\[\begin{align*} \lim_{x\to 6^{+}} h(x) &= \lim_{x\to 6^+} x^2-5x-6\\ \\ &= 6^2-5(6)-6 && \text{plug in $x=6$ (Property 7)}\\ \\ &= 36-30-6 && \text{simplify}\\ \\ &= 0 \end{align*}\]

Step 3:   Check to see if the two limits are equal.

Since \(\displaystyle \lim_{x\to 6^-} h(x) = 0\) and \(\displaystyle \lim_{x\to 6^+} h(x) = 0\), we conclude that the limit exists and

\[\lim_{x\to 6} h(x) = 0.\]

Warning

Keep in mind that just because \(\displaystyle\lim_{x\to 6} h(x) \neq h(6) = 8\), this does not mean that the limit does not exist.