Chapter 11: Asymptotes

Horizontal Asymptotes


y=by=b is a HORIZONTAL ASYMPTOTE of f(x)f(x) if limxf(x)=b\lim_{x \rightarrow \infty} f(x) =b  or limxf(x)=b\lim_{x \rightarrow - \infty} f(x) =b 


Find the horizontal asymptotes of y=x22x2+1y = \frac{x^2}{2x^2+1}

We find the limits at infinity

limxx22x2+1 \lim_{x \rightarrow \infty} \frac{x^2}{ 2x^2 +1}

== limxx22x2+1(1/x21/x2)\lim_{x \rightarrow \infty} \frac {x^2}{2x^2 +1 } (\frac{1/x^2 }{1/x^2})  \rightarrow  Cancel x2x^2

=limx12+1x2= \lim_{x \rightarrow \infty} \frac{1}{2 + \frac{1}{x^2}} 

=12+0 = \frac {1}{2+0 } 


Thus, y=12y= \frac{1}{2} is a horizontal asymptote

* \frac {\infty}{\infty} is tricky. It can equal any finite value


Find the horizontal asymptotes of y=3x3+5x10x33y= \frac{3x^3 +5x}{10x^3 -3}

  • Cancel x3x^3
  • Visually, the graph flattens


Find the horizontal asymptote of y=exy=e^x

Limits to infinity

limxex= \lim_{x \rightarrow \infty} e^x =\infty  and limxex=0\lim_{x \rightarrow \infty} e^x =0

Observe that ex0e^ x \geq 0 


How does y=x3x4+1y = \frac {x^3} {x^4+1 }  approach the line y=0y=0?

limxx3x4+1\lim_{ x \rightarrow \infty} \frac{x^3}{x^4+1}=++= \frac{+}{+}

Thus , limxx3x4+1=0+\lim_{ x \rightarrow \infty} \frac{x^3}{x^4+1} = 0^+

limxx3x4+1=+\lim_{x \rightarrow - \infty} \frac{x^3}{x^4+1}=\frac{-}{+}

Thus, limxx3x4+1=0\lim_{x \rightarrow \infty} \frac {x^3}{x^4+1 }= 0^-

The graph crosses y=0y=0

This graph was made using a graphing calculator

Why are there bumps?


How does y=x56x53y= \frac {x^5}{6x^5 -3 }  approaches y=16y = \frac{1}{6}?

limxx56x53=++\lim_{x \rightarrow \infty} \frac{x^5 }{6x ^5-3 } = \frac{+}{+ }

Thus, limxf(x)=16+\lim_{x \rightarrow \infty} f(x) = \frac {1}{6} ^+

limxx56x53=\lim_{x \rightarrow \infty} \frac{ x^5}{6x^5 - 3} = \frac{-}{-}

Thus limxf(x)=16+\lim_{x \rightarrow - \infty} f(x) = \frac {1}{6} ^+

Vertical Asymptotes


Where are the vertical asymptotes of y=x56x53y= \frac{x^5}{6x^5-3}?

Find zeroes of denominator

y=x56x53=x56(x512)y= \frac{x^5}{6x^5-3} = \frac {x^5}{6(x^5- \frac{1}{2})}

6(x512)=0x512=06(x^5-\frac{1}{2}) = 0 \Leftrightarrow x^5- \frac{1}{2} =0

x5=12x=125=C\Leftrightarrow x^5 = \frac {1}{2} \Leftrightarrow x = \frac{1}{ \sqrt[5]{2}} =C

Find limit to x=125x =\frac{1}{ \sqrt[5]{2}}

limxc+x56x53=++=+\lim_{x \rightarrow c^+} \frac{ x^5 }{6x^5-3 }= \frac{+}{+} = + \infty  x>1256x53>0x> \frac{1}{\sqrt[5]{2}} \Leftrightarrow 6x^5-3 >0

limxcx56x53=+=\lim_{x \rightarrow c^-} \frac{ x^5 }{6x^5-3 }= \frac{+}{-} = - \infty  x<1256x53<0x < \frac{1}{\sqrt[5]{2}} \Leftrightarrow 6x^5-3 <0

Thus, limxc+f(x)=\lim_{x \rightarrow c^+} f(x) = \infty  and limxcf(x)=\lim_{x \rightarrow c^-} f(x) = - \infty


What is the slope of this graph near asymptotes?

  • Flattens out near horizontal.
  • Spike near vertical
  • Changes sign


Horizontal Asymptotes

  • y=blimxf(x)=by =b \Leftrightarrow \lim_{x \rightarrow \infty} f(x) = b or limxf(x)=b\lim_{x \rightarrow - \infty} f(x)=b
  • 0, 1, or 2 horizontal asymptotes
  • The graph flattens out.
  • Approach from above (y=b+)(y=b^+) or below (y=b)(y=b^-)

Vertical Asymptotes

  • x=climxc+f(x)=±orlimxcf(x)=±x=c \Leftrightarrow \lim_{x \rightarrow c^+} f(x) = \pm \infty or \lim_{x \rightarrow c^-} f(x) = \pm \infty 
  • x=climxc+f(x)=±x=c \Leftrightarrow \lim_{x \rightarrow c^+} f(x) = \pm \infty or limxcf(x)=±\lim_{x \rightarrow c^-} f(x) = \pm \infty 
  • Any number of vertical asymptotes
  • The graph has a spike
  • The graph may change sign


Classify all asymptotes of y=x35x35xy = \frac{x^3}{5x^3-5x}


Classify all asymptotes of y=sin(x)xy = \frac{ \sin(x)}{ x}

Given the asymptotes how do we fill in the rest of the graph?

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