Tends to infinity

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Posts Tagged ‘Calculus’


Posted by tendstoinfinity on December 2, 2010

While teaching curve tracing I always found if the students doesn’t have the idea of asymptotes they are not able to understand the whole idea of curve tracing that is why this topic

Asymptotes: By def., asymptote is that st. line which meets the curve at two points at infinity. Now the method to find the equation of asymptote

Method 1:

  • If at least one of  lim_{x\mapsto a+}f(x)=\infty  or lim_{x\mapsto a-}f(x)=\infty  then x = a is vertical asymptote. The curve lies on the right or left of the asymptote according as the case.
  • If  lim_{x\mapsto\pm\infty}f(x)=A then y = A is horizontal asymptote.
  • If lim_{x\mapsto\pm\infty}f(\dfrac{y}{x})=m and lim_{x\mapsto\pm\infty}f(x)-mx=c exist then y = mx + c is an asymptote.
  1. Example find the asymptote of y=xe^{\frac{1}{x}}

It is clear that lim_{x\mapsto 0+} y=lim_{x\mapsto 0+} (xe^{\frac{1}{x}}) =+\infty (by using L’hospital rule)                         (what about lim_{x\mapsto 0-}xe^{\frac{1}{x}} ?)

Hence x = 0 is vertical asymptote.

There is no vertical asymptotes

Again for oblique asymptote

m= lim_{x\mapsto \infty}(\dfrac{y}{x})=lim_{x\mapsto \infty}e^{\dfrac{1}{x}}=1 and

c=lim_{x\mapsto \pm\infty}(y-mx)=lim_{x\mapsto \pm\infty}(xe^{\frac{1}{x}}-x)=lim_{x\mapsto \pm\infty}(x(e^{\frac{1}{x}}-1))=1 (expanding e^{\dfrac{1}{x}} )

hence y =x + 1 is an asymptote.

What about the annexed graph ?

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