Limits as x tends to a

The function

\[f(x)=\frac{x^2+4x-12}{x^2-2x} \]

is not defined when \(x=2\), it does however have a limit as \(x\rightarrow 2\), the limit is 4. We write this as

\[\lim_{x\rightarrow 2}\frac{x^2+4x-12}{x^2-2x} = 4 \]

In this example it is easy to find the limit without graphing it

\[\lim_{x\rightarrow 2}\frac{x^2+4x-12}{x^2-2x} = \lim_{x\rightarrow 2} \frac{(x-2)(x+6)}{x(x-2)} = \lim_{x\rightarrow 2} \frac{x+6}{x} =\frac{2+6}{2}=4\]

Infinity as a limit, vertical asymptotes

The expression \(1/x^2\) gets larger and larger as \(x\) gets closer to zero. We write this as:

\[\lim_{x\rightarrow 0}\frac{1}{x^2}=\infty\]

Even though the notation above is often used, infinity is not usually seen as a limit since the infinity is not a number (unless you define it to be a number, see Extended real number line).

The graph of the function \(f(x)=1/x^2\) gets closer and closer to the line \(x=0\) as \(x\rightarrow 0\), the line \(x=0\) is a vertical asymptote.

Note that this is not true for general functions, see exercise 1 below.

Let

\[f(x)=\frac{x^2-3}{x^2+2x-8} \]

The denominator is zero when \(x=2\) and when \(x=-4\). The function has two vertical asymptotes.

We can approach the value \(x=2\) from two directions, either \(x\lt 2\) or \(2\lt x\) . In order for a limit to exist, the limits we get from the two directions must be the same. In this case a limit does not exist, not even infinity. The function does however have a vertical asymptote. Sometimes we want to specify that an expression has different limits depending on whether we approach the limit from the left or the right. We use a + (\(x\) is greater than) or a - (\(x\) is less then), to distinguish between the two cases.

\[\lim_{x\rightarrow 2^+} \frac{x^2-3}{x^2+2x-8} = \infty \]

\[\lim_{x\rightarrow 2^-} \frac{x^2-3}{x^2+2x-8} = -\infty \]

Difficult limits

In some cases you use common sense to find limits:

\[\frac{1}\infty = 0 \hspace{1 cm} \frac{1}{0}=\infty \hspace{1 cm} 1+\infty=\infty \hspace{1 cm} 2\cdot \infty = \infty\]

(Don't write like this on ↑ the exam)

It is difficult in some cases:

\[\frac{\infty}\infty = ? \hspace{1 cm} \frac{0}{0}=? \hspace{1 cm} \infty-\infty=? \hspace{1 cm} 0\cdot \infty = ?\]

Exercises

Graph the following functions to find the limits, if they exist.

1. \( \displaystyle{\lim_{x\rightarrow 0}\sin\left( \frac{1}{x}\right)} \)
2. \( \displaystyle{\lim_{x\rightarrow 0} x \cdot \sin\left( \frac{1}{x}\right)} \)
3. \( \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x}{x} } \)
4. \( \displaystyle{\lim_{x\rightarrow 0} \frac{\cos x -1}{x} } \)