Computing Limits Algebraically

Example 1

The limit of a constant function

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

Step 1:   Recall the limit property of a constant function.

For any real number \(c\), \(\lim\limits_{x\to a}c = c\).

See Properties of Finite Limits for a list of all limit properties.

Step 2:   Apply the limit property of a constant function.

Since \(f(x) = 2\) is a constant function, \(\lim\limits_{x\to 1}f(x) = 2\), which implies that the one-side limit also exists and

\[\lim\limits_{x\to 1^{-}}f(x) = 2.\]

Example 2

The limit of a rational function

Evaluate \(\lim\limits_{x\to -1} \dfrac{x^3 +4x^2-x-3}{2x^2+1}\).

Step 1:   Recall the limit property of a rational function.

If \(f(x)\) is a rational function and \(a\) is in the domain of \(f(x)\), then \(\lim\limits_{x\to a}f(x) = f(a)\).

See Properties of Finite Limits for a list of all limit properties.

Step 2:   Determine if \(x=-1\) is in the domain of the function.

The function

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

is a rational function and since \(2(-1)^2 + 1 \neq 0\) (i.e., the denominator is not equal to zero when \(x=-1\)), it follows that \(-1\) is in the domain of \(g\).

Therefore, the limit exists and

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

Example 3

The limit of a rational function

Evaluate \(\lim\limits_{x\to 2} \dfrac{x^2 +4x-1}{x-2}\).

Step 1:   Determine if \(x=2\) is in the domain of the function.

The function

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

is a rational function and since \(2-2 = 0\) (i.e., the denominator equals zero when \(x=2\)), it follows that \(2\) is not in the domain of \(h\). Unfortunately, this means that the limit property we used in the previous example does not apply.

Step 2:   Determine if the limit exists.

When we evaluate the numerator of \(h(x)\) at \(x=2\), we get

\[2^2 +4(2) - 1 = 11.\]

Since the denominator goes to zero but the numerator does not, we can immediately conclude that \(\lim\limits_{x\to 2} \dfrac{x^2 +4x-1}{x-2}\) does not exist.

Moreover, if we let \(x\) approach \(2\) from the left (i.e., \(x<2\)), the values of \((x^2 +4x-1)/(x-2)\) will be negative and we can conclude that

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

And if we let \(x\) approach \(2\) from the right (i.e., \(x>2\)), the values of \((x^2 +4x-1)/(x-2)\) will be positive and we can conclude that

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

Even though we can say that the one-sided limits are equal to positive or negative infinity, this still means that the one-sided limits do not exist, and therefore the original limit does not exist either.

Evaluating the Limit of a Rational Function

When evaluating the limit of a rational function, \(\lim\limits_{x\to a} \dfrac{f(x)}{g(x)}\), start by plugging in \(x=a\) into both \(f(x)\) and \(g(x)\).

• If \(g(a) \neq 0\), then the limit exists and is equal to \(f(a)/g(a)\).

• If \(g(a)=0\) and \(f(a)\neq 0\), then the limit does not exist. By carefully analyzing the sign of the numerator and of the denominator, we can determine if the one-sided limits go to positive or negative infinity.

• If \(g(a) = 0\) and \(f(a)=0\), it’s still possible the limit exists. We will consider this situation in the next section.

Example 4

The limit of a piecewise 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:   Recall the limit property of a polynomial

If \(f(x)\) is a polynomial function, then \(\lim\limits_{x\to a}f(x) = f(a)\).

See Properties of Finite Limits for a list of all limit properties.

Step 2:   Compute the limit from the left.

Since we are approaching 6 from the left, we can assume that \(x<6\), which means \(h(x) = x^2-4x-12\) and we can apply the limit property of a polynomial.

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

Step 3:   Compute the limit from the right.

Since we are approaching 6 from the right, we can assume that \(x>6\), which means \(h(x) = x^2-5x-6\), and we can apply the limit property of a polynomial.

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

Step 4:   Check to see if the two one-sided limits are equal.

Since \(\displaystyle \lim_{x\to 6^-} h(x) = 0\) and \(\displaystyle \lim_{x\to 6^+} h(x) = 0\) (i.e., the two one-sided limits both exist and are equal), we conclude that the limit exists and

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

Warning

Notice that we did not use the fact that \(h(6) = 8\) while computing \(\displaystyle\lim_{x\to 6} h(x)\). Futhermore, the fact that \(\displaystyle\lim_{x\to 6} h(x) \neq h(6)\) does not mean that the limit does not exist.

Example 5

The limit of a piecewise function

Evaluate \(\displaystyle\lim_{x\to 4} g(x)\), where

\[\begin{equation*} g(x)= \begin{cases} \dfrac{x+5}{3} & \hbox{if } x\leq 4 \\ 6 - x^2& \hbox{if } x>4 \end{cases} \end{equation*}\]

Step 1:   Compute the limit from the left.

Since we are approaching 4 from the left, we can assume that \(x<4\), which means \(g(x) =\frac{x+5}{3}\) and we can apply the limit property of a polynomial.

\[\begin{align*} \lim_{x\to 4^{-}} g(x) &= \lim_{x\to 4^-} \frac{x+5}{3} \\ \\ &= \frac{4+5}{3} && \text{plug in $x=4$}\\ \\ &= 9/3 && \text{simplify}\\ \\ &= 3 \end{align*}\]

Step 2:   Compute the limit from the right.

Since we are approaching 4 from the right, we can assume that \(x>4\), which means \(g(x) = 6 - x^2\), and we can apply the limit property of a polynomial.

\[\begin{align*} \lim_{x\to 4^{+}} g(x) &= \lim_{x\to 6^+} 6 - x^2\\ \\ &= 6 - 4^2 && \text{plug in $x=4$}\\ \\ &= 6 - 16 && \text{simplify}\\ \\ &= -10 \end{align*}\]

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

Since \(\displaystyle \lim_{x\to 4^-} g(x) = 3\) and \(\displaystyle \lim_{x\to 4^+} g(x) = -10\) (i.e., the two one-sided limits both exist but are not equal), we conclude that \(\lim\limits_{x\to 4} g(x)\) does not exist.