How to Find Infinite Limits and Vertical Asymptotes?

A vertical asymptote is a place where the function is not defined and the limit of the function does not exist. This is because as \(1\) approaches the asymptote even small shifts in the \(x\)-value lead to arbitrarily large fluctuations in the value of the function.

Related Topics

• Connecting Limits at Infinity and Horizontal Asymptotes

A step-by-step guide to infinite limits and vertical asymptotes

If the values of \(f(x)\) become very big positive numbers (or very large negative numbers) as \(x\) approaches \(a\) from the left, we declare:

\(lim_{x\to a^-} f(x)= ∞\:\ or\:\ (-∞)\)

If \(x\) approaches \(a\) from the right, we say:

\(lim_{x\to a^+} f(x)= ∞\:\ or\:\ (-∞)\)

If both one-sided limits have the same behavior, we say:

\(lim_{x\to a} f(x)= ∞\:\ or\:\ (-∞)\)

The graph \(y = f(x)\) also includes a vertical asymptote at \(x = a\) in these circumstances.

Steps for determining vertical asymptotes given equations:

1. Factor the equation to make it simple.
2. Find the denominator that equals zero.
3. If you’re looking for something different (for limit problems), plug-in numbers relatively near to the left and right of each value to determine the sign (positive or negative). This indicates whether the left- or right-handed boundaries are positive or negative infinity.

Infinite Limits and Vertical Asymptotes – Example 1:

Find the vertical asymptotes of the graph \(f(x)=\frac{2x^2+5}{-x^2-x+2}\).

First, factor the numerator and denominator:

\(f(x)=\frac{2x^2+5}{-x^2-x+2}\) \(=\frac{2x^2+5}{-(x-1)(x+2)}\)

To find vertical asymptotes, set the denominator to zero to determine where this function is not defined:

\(-(x-1)(x+2)=0\)

\((-x+1)=0 → -x=-1→ x=1\)

\((-x-2)=0 → -x=2 → x=-2\)

Neither \(x=-2\) nor \(x=1\) are zeros of the numerator, so the two values indicate two vertical asymptotes.

Infinite Limits and Vertical Asymptotes – Example 2:

Find the vertical asymptotes of the graph \(f(x)=\frac{x-2}{x^2-4}\).

First, factor the numerator and denominator:

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

There is a factor in the denominator that is not in the numerator, \((x+2)\). The zero for this factor is :

\((x+2)=0 → x=-2\)

The vertical asymptote is \(x=-2\).

Infinite Limits and Vertical Asymptotes – Example 3:

Find the value of \(lim _{x\to \infty }\left(\frac{2x^2+3x}{10x^2+x}\right)\).

For infinity limits, the leading term must be considered in both the numerator and the denominator. Here, we have the case that the exponents are equal in the leading expressions. Therefore, the limit at infinity is simply the ratio of the coefficients of the leading expressions.

\(lim _{x\to \infty }\left(\frac{2x^2+3x}{10x^2+x}\right)\) \(=\frac{2x^2}{10x^2}=\frac{2}{10}=\frac{1}{5}\)

Exercises for Infinite Limits and Vertical Asymptotes

Find the vertical asymptotes of the functions and the value of each limit.

1. \(\color{blue}{f(x)=\frac{-2x^2+2}{x^3-6x^2+5x}}\)
2. \(\color{blue}{f(x)=\frac{2x^2+7x+12}{2x^2+5x-12}}\)
3. \(\color{blue}{f(x)=\frac{3x^2-9}{x^2+7x+12}}\)
4. \(\color{blue}{lim _{x\to 4^-}\left(\frac{3}{x-4}\right)}\)
5. \(\color{blue}{lim _{x\to 2^+}\left(\frac{x^2-2x-3}{x^2-x-2}\right)}\)
6. \(\color{blue}{lim _{x\to -\infty }\left(\frac{4x^5-3x^2+2}{2x^3-x^2+1}\right)}\)

1. \(\color{blue}{x=0 , x=5}\)
2. \(\color{blue}{x=-4, x=\frac{3}{2}}\)
3. \(\color{blue}{x=-3, x=-4}\)
4. \(\color{blue}{-\infty \:}\)
5. \(\color{blue}{-\infty \:}\)
6. \(\color{blue}{\infty \:}\)