If \(f(x)\) approaches a real number \(L\) as \(x\) approaches \(a\) from the left \((x<a)\), then
\(\lim_{x\to a^-}f(x)=L\) (left-handed one sided limit)
Likewise, if \(f(x)\) approaches a real number \(M\) as \(x\) approaches \(a\) from the right \((x>a)\), then
\(\lim_{x\to a^+}f(x)=M\) (right-handed one-sided limit)
Existence of a Limit of a Function
Theorem.
A function \(f(x)\) has a limit as approaches a number \(a\) if and only if its left-handed and right-handed limits that exist and are equal:
\(\lim_{x\to a}f(x)=L\quad\text{if and only if}\quad\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L\)
Example.
For the function \(f(x)= \begin{cases} x^3-3x, & \quad\mathrm{if}-1\leq x<1 \\ x-5, & \quad\mathrm{if}x\geq1 & & \end{cases}\), evaluate \(\lim_{x\to1^+}f(x)\) and \(\lim_{x\to1^-}f(x)\). Does the limit \(\lim_{x\to1}f(x)\) exist?
Solution.
Since \(f(x)=x-5\) for \(x\geq1,\)
\(\lim_{x\to1^+}(x-5)=1-5=-4\)
\(\mathrm{Since~}f(x)=x^3-3x\mathrm{~for~}-1\leq x<1,\)
\(\lim_{x\to1^-}x^3-3x=\lim_{x\to1^-}(1)^3-3(1)=1-3=-2\)
\(\mathrm{since~}\lim_{x\to1^-}f(x)=-2\neq-4=\lim_{x\to1^+}f(x),\)
\(\lim_{x\to1}f(x)\text{ does not exist.}\)
Infinite Limits
\(\text{If}\lim_{x\to a^+}f(x)=\pm\infty\quad\mathrm{or}\quad\lim_{x\to a^-}f(x)=\pm\infty,\text{then it is said that}\lim_{x\to a}f(x)\) does not exist.
Vertical Asymptotes
The line \(x=a\) is a vertical asymptote of the graph of \(y=f(x)\) if either \(\lim_{x\to a^+}f(x)=\pm\infty\quad\mathrm{or}\quad\lim_{x\to a^-}f(x)=\pm\infty.\)
Geometrically, the graph of a function \(f(x)\) is said to have a vertical asymptote at \(x=a\) if \(f(x)\) increases or decreases without bound as \(x\) approaches \(a\), from either the right or the left, or from both directions.
In general, a rational function \(R(x)=\frac{p(x)}{q(x)}\) has a vertical asymptote \(x=a\) whenever \(q(a)=0,\mathrm{but~}p(a)\neq0.\)
Example.
Find \(\lim_{x\to3^+}\frac2{x-3}\) and \(\lim_{x\to3^-}\frac2{x-3}.\) Is there a vertical asymptote?
Solution.
When \(x=3\), the denominator in \(f(x)=\frac2{x-3}\) is 0 and the numerator is a nonzero number. Thus, the limit of \(f(x)\) as \(x\) approaches 3 is either \(\infty\mathrm{~or~}-\infty.\)
Consider the right-handed limit \(\lim_{x\to3^+}\frac2{x-3}\).
Since \(x>3\), the denominator \(x-3\) is positive, and thus, the fraction \(\frac2{x-3}\) is positive.
Therefore, \(\lim_{x\to3^+}\frac2{x-3}=+\infty \), and the function \(f(x)=\frac2{x-3}\) has a vertical asymptote at \(x=3\).
Compute the left-handed limit \(\lim_{x\to3^-}\frac2{x-3}\) in the same way. Alternatively, pick a value of \(x\) very close to the left of 3, say \(x=2.9\), and substitute this value into \(f(x)=\frac2{x-3}\). A negative number is obtained. It follows that \(\lim_{x\to3^-}\frac2{x-3}=-\infty.\) Thus, the function \(f(x)=\frac2{x-3}\) has a vertical asymptote at \(x=3\).
As a result, the function \(f(x)=\frac2{x-3}\) has a vertical asymptote at \(x=3\) from both sides.