# Tends to infinity

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# Posts Tagged ‘Calculus’

## ASYMPTOTE

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 ?