Functions: Limits and asymptotes
Oblique asymptotes
In addition to horizontal and vertical asymptotes, there is a third type of asymptote, the oblique asymptote. Another name for the oblique asymptote is the slant asymptote. This is an oblique line of the form #y=ax+b#, that gets approached by the graph of the function when #x# gets really big or small.
Oblique asymptotes
The function #\blue{f(x)}# has an oblique asymptote #y=\green{ax+b}# to infinity if #\blue{f(x)}# can be rewritten as \[\blue{f(x)}=\green{ax+b}+\orange{g(x)}\] where
\[\lim\limits_{x \to \infty}\orange{g(x)}=0 \text{ and } a\ne 0\]
In a similar way #\blue{f(x)}# has an oblique asymptote #y=\green{ax+b}# to minus infinity if #\blue{f(x)}# can be rewritten as \[\blue{f(x)}=\green{ax+b}+\orange{g(x)}\] where
\[\lim\limits_{x \to -\infty}\orange{g(x)}=0 \text{ and } a\ne 0\]
Example
#\blue{f(x)}=\blue{\frac{x^2+1}{x}}#
has an oblique asymptote #y=\green{x}#
to #\infty# and #-\infty#, since
#\blue{\frac{x^2+1}{x}}=\green{x}+\orange{\frac{1}{x}}#
where
#\lim_{x \to \infty}\orange{\frac{1}{x}}=0#
and
#\lim_{x \to -\infty}\orange{\frac{1}{x}}=0#
\[f(x)={{x^2+3 x-4}\over{x+5}}\]
Determine the oblique asymptotes of this function. Give your answer in the form "#y=a\cdot x +b#" with the correct numbers for #a# and #b#. If there are multiple oblique asymptotes, then click the plus sign to add an answer field. If there are no oblique asymptotes, enter " none".
The degree of the numerator is #2# and the degree of the denominator is #1#. The degree of the numerator is exactly one greater than the denominator so that the function has an oblique asymptote.
We can find the olique asymptote by dividing the numerator by the denominator using long division:
\[\begin{array}{rcl}
\displaystyle f(x)
&=& \displaystyle {{x^2+3 x-4}\over{x+5}}\\
&& \qquad \blue{\text{substituted}}\\
&=& \displaystyle x-2 + {{6}\over{x+5}} \\
&& \qquad \blue{\text{long division}}\\
\end{array}\]
We see that #y=x-2# is a candidate for an oblique asymptote. We still have to check if the last term goes to #0# if #x\to \pm\infty# :
\[\displaystyle\lim_{x\to\infty}{{6}\over{x+5}}=0\]
\[\displaystyle\lim_{x\to-\infty}{{6}\over{x+5}}=0\]
The last term indeed goes to #0# if #x\to \pm\infty#, so the oblique asymptote at positive and negative infinity is #y=x-2#
\[f(x)={{x^2+3 x-4}\over{x+5}} = \frac{\left(x-1\right)\cdot \left(x+4\right)}{x+5}\]
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