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Asymptotes

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Definition. The line x=ax=a is called vertical asymptote of the curve y=f(x)y=f{{\left({x}\right)}} if at least one of the following statements is true:

  1. lim⁡x→af(x)=∞\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}=\infty
  2. lim⁡x→a+f(x)=∞\lim_{{{x}\to{{a}}^{+}}}{f{{\left({x}\right)}}}=\infty
  3. lim⁡x→a−f(x)=∞\lim_{{{x}\to{{a}}^{{-}}}}{f{{\left({x}\right)}}}=\infty
  4. lim⁡x→af(x)=−∞\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}=-\infty
  5. lim⁡x→a+f(x)=−∞\lim_{{{x}\to{{a}}^{+}}}{f{{\left({x}\right)}}}=-\infty
  6. lim⁡x→a−f(x)=−∞\lim_{{{x}\to{{a}}^{{-}}}}{f{{\left({x}\right)}}}=-\infty

For example y-axis (x=0)\left({x}={0}\right) is a vertical asymptote of the curve y=1x2{y}=\frac{{1}}{{{x}}^{{2}}} because lim⁡x→01x2=∞\lim_{{{x}\to{0}}}\frac{{1}}{{{x}}^{{2}}}=\infty.

Geometrically vertical asymptote is vertical line to which approaches graph of the function but never intersects it. As function approaches a{a} values of function become so large, that we can't draw them.

vertical asymptoteExample 1. Find vertical asymptotes of f(x)=xx−1{f{{\left({x}\right)}}}=\frac{{x}}{{{x}-{1}}}.

In Limits Involving Infinity note we saw that if lim⁡x→a+f(x)=L≠0\lim_{{{x}\to{{a}}^{+}}}{f{{\left({x}\right)}}}={L}\ne{0} and lim⁡x→a+g(x)=0\lim_{{{x}\to{a}+}}{g{{\left({x}\right)}}}={0} then lim⁡x→a+(f(x)g(x))=∞\lim_{{{x}\to{{a}}^{+}}}{\left(\frac{{f{{\left({x}\right)}}}}{{g{{\left({x}\right)}}}}\right)}=\infty (Law 8 for one-sided limits).

So, to find points where function approach infinity, we need to find points where denominator equals 0, and make sure that numerator doesn't equal 0 at this point.

The only point where denominator equals 0 is point 1, and numerator doesn't equal 0 at this point.

Let's find out one-sided limits at this point. When x{x} approaches 1 from the right then x−3{x}-{3} becomes very small positive value and x{x} approaches 1. Thus, lim⁡x→1+xx−1=∞\lim_{{{x}\to{{1}}^{+}}}\frac{{{x}}}{{{x}-{1}}}=\infty When x{x} approaches 1 from the left then x−1{x}-{1} becomes very small negative value and x{x} approaches 1. Thus, lim⁡x→1−xx−1=−∞\lim_{{{x}\to{{1}}^{{-}}}}\frac{{{x}}}{{{x}-{1}}}=-\infty. So, we showed that both one-sided limits are infinite. In fact it was sufficiently to show that at least one of the limits is infinite.

Thus, x=1{x}={1} is vertical asympote.

In fact function can have more than vertical asymptote.

Example 2. Find vertical asymptotes of y=tan⁡(x){y}={\tan{{\left({x}\right)}}}.more than one vertical asymptote

Recall that by definition tan⁡(x)=sin⁡(x)cos⁡(x){\tan{{\left({x}\right)}}}=\frac{{{\sin{{\left({x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}.

This means that there will be vertical asymptotes at points where cos⁡(x)=0{\cos{{\left({x}\right)}}}={0}, i.e. points of the form x=π2+πk,k∈Z{x}=\frac{\pi}{{2}}+\pi{k},{k}\in\mathbb{Z}.

Therefore, tan⁡(x){\tan{{\left({x}\right)}}} has infinitely many asymptotes.

Example 3. Find vertical asymptotes of f(x)=x2−xx2+x−2{f{{\left({x}\right)}}}=\frac{{{{x}}^{{2}}-{x}}}{{{{x}}^{{2}}+{x}-{2}}}.

vertical asymptote exampleSince x2+x−2=(x+2)(x−1){{x}}^{{2}}+{x}-{2}={\left({x}+{2}\right)}{\left({x}-{1}\right)} then denominator equals 0 when x=−2{x}=-{2} and x=1{x}={1}.

However, at point x=1{x}={1} numerator also equals 0. We need to additionally check this point.So, lim⁡x→1x2−xx2+x−2=lim⁡x→1x(x−1)(x+2)(x−1)=\lim_{{{x}\to{1}}}\frac{{{{x}}^{{2}}-{x}}}{{{{x}}^{{2}}+{x}-{2}}}=\lim_{{{x}\to{1}}}\frac{{{x}{\left({x}-{1}\right)}}}{{{\left({x}+{2}\right)}{\left({x}-{1}\right)}}}=

=lim⁡x→1xx+2=11+2=13=\lim_{{{x}\to{1}}}\frac{{x}}{{{x}+{2}}}=\frac{{1}}{{{1}+{2}}}=\frac{{1}}{{3}}.

Thus, there is only one vertical asymptote x=−2{x}=-{2}. This example showed, that you should be very cautious: you should check whether numerator equals 0 at points where denominator equals 0.

Definition. Line y=L{y}={L} is called horizontal asymptote of the function f(x){f{{\left({x}\right)}}} if either lim⁡x→∞f(x)=L\lim_{{{x}\to\infty}}{f{{\left({x}\right)}}}={L} or lim⁡x→−∞f(x)=L\lim_{{{x}\to-\infty}}{f{{\left({x}\right)}}}={L}.

Example 4. Find horizontal asymptotes of y=1+1x{y}={1}+\frac{{1}}{{x}}.horizontal asymptote

Since lim⁡x→∞(1+1x)=1\lim_{{{x}\to\infty}}{\left({1}+\frac{{1}}{{x}}\right)}={1} then line y=1{y}={1} is horizontal asymptote.

In fact also lim⁡x→−∞(1+1x)=1\lim_{{{x}\to-\infty}}{\left({1}+\frac{{1}}{{x}}\right)}={1}, but it suffices one limit to state that y=1{y}={1} is horizontal asymptote.

In general function can have more than one horizontal asymptote.

Example 5. Find horizontal asymptotes of y=arctan⁡(x){y}={\operatorname{arctan}{{\left({x}\right)}}}.more than one horizontal asymptote

Since lim⁡x→∞arctan⁡(x)=π2\lim_{{{x}\to\infty}}{\operatorname{arctan}{{\left({x}\right)}}}=\frac{\pi}{{2}} and lim⁡x→−∞arctan⁡(x)=−π2\lim_{{{x}\to-\infty}}{\operatorname{arctan}{{\left({x}\right)}}}=-\frac{\pi}{{2}} then there are two vertical asymptotes: y=π2{y}=\frac{\pi}{{2}} and y=−π2{y}=-\frac{\pi}{{2}}.



Definition. Line y=mx+b{y}={m}{x}+{b} is called slant (oblique) asymptote of function y=f(x){y}={f{{\left({x}\right)}}} if either lim⁡x→∞(f(x)−(mx+b))=0\lim_{{{x}\to\infty}}{\left({f{{\left({x}\right)}}}-{\left({m}{x}+{b}\right)}\right)}={0} or lim⁡x→−∞(f(x)−(mx+b))=0\lim_{{{x}\to-\infty}}{\left({f{{\left({x}\right)}}}-{\left({m}{x}+{b}\right)}\right)}={0}.

We calculate m{m} as follows: m=lim⁡x→∞f(x)x(m=lim⁡x→−∞f(x)x){m}=\lim_{{{x}\to\infty}}\frac{{{f{{\left({x}\right)}}}}}{{x}}{\left({m}=\lim_{{{x}\to-\infty}}\frac{{{f{{\left({x}\right)}}}}}{{x}}\right)}.

If this limit is not finite or doesn't exist then there is no oblique asymptote.

b{b} is calculated as follows: b=lim⁡x→∞(f(x)−mx)(b=lim⁡x→−∞(f(x)−mx)){b}=\lim_{{{x}\to\infty}}{\left({f{{\left({x}\right)}}}-{m}{x}\right)}{\left({b}=\lim_{{{x}\to-\infty}}{\left({f{{\left({x}\right)}}}-{m}{x}\right)}\right)}.

Example 6. Find all asymptotes of y=x+1x{y}={x}+\frac{{1}}{{x}}.slant asymptote

There are no horizontal asymptotes because lim⁡x→∞(x+1x)=∞\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}=\infty and lim⁡x→−∞(x+1x)=−∞\lim_{{{x}\to-\infty}}{\left({x}+\frac{{1}}{{x}}\right)}=-\infty.

There is vertical asymptote x=0{x}={0} because lim⁡x→0+(x+1x)=∞\lim_{{{x}\to{{0}}^{+}}}{\left({x}+\frac{{1}}{{x}}\right)}=\infty.

This function also has oblique asymptote. Indeed, m=lim⁡x→∞x+1xx=lim⁡x→∞(1+1x2)=1{m}=\lim_{{{x}\to\infty}}\frac{{{x}+\frac{{1}}{{x}}}}{{x}}=\lim_{{{x}\to\infty}}{\left({1}+\frac{{1}}{{{x}}^{{2}}}\right)}={1}.

So, m=1{m}={1}. Now, b=lim⁡x→∞(x+1x−1⋅x)=lim⁡x→∞1x=0{b}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}-{1}\cdot{x}\right)}=\lim_{{{x}\to\infty}}\frac{{1}}{{x}}={0}. So, b=0{b}={0}.

Therefore, slant asymptote is y=x{y}={x}. On the figure to the right blue line is vertical asymptote x=0{x}={0}, green line is slant asymptote y=x{y}={x}.

Note, that we will obtain same slant asymptote when x→−∞{x}\to-\infty.

In general, it is a good practice to treat cases x→∞{x}\to\infty and x→−∞{x}\to-\infty separately.