Equation of Infinity

[Jason76]

Find the equations of the vertical asymptote, given only the limit.

\(\displaystyle \lim x \to -7\) of \(\displaystyle f(x)\) Equation \(\displaystyle = \dfrac{x}{0} = \dfrac{-7}{0} = -\infty\)

\(\displaystyle \lim x \to 6-\) of \(\displaystyle f(x)\) Equation \(\displaystyle = \dfrac{x}{0-} = \dfrac{6}{0-}= -\infty\)

\(\displaystyle \lim x \to 6+\) of \(\displaystyle f(x)\) Equation \(\displaystyle = \dfrac{x}{0+} = \dfrac{6}{0+} = \infty\)

Is this right

 

[daon2]

The equation of a vertical line is of the form \(\displaystyle x=a\), where in your case \(\displaystyle a\) is the location of an asymptotic discontinuity.

 

[Jason76]

The equation of a vertical line is of the form \(\displaystyle x=a\), where in your case \(\displaystyle a\) is the location of an asymptotic discontinuity.

So what would be a hint on this one?

\(\displaystyle \lim x \to -7\) of \(\displaystyle f(x)\) Equation \(\displaystyle = \dfrac{x}{0} = \dfrac{-7}{0} = -\infty\)

 

[daon2]

So what would be a hint on this one?

\(\displaystyle \lim x \to -7\) of \(\displaystyle f(x)\) Equation \(\displaystyle = \dfrac{x}{0} = \dfrac{-7}{0} = -\infty\)

x=-7

 

[Jason76]

x=-7

So simple.

 

[HallsofIvy]

\(\displaystyle = \dfrac{x}{0} = \dfrac{-7}{0} = -\infty\)

This is not correct. These fractions are NOT "equal to infinity"- they do not exist at all- you cannot divide by 0! (What is true is that \(\displaystyle \lim_{h\to 0} \dfrac{-7}{h}= -\infty\)).