(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]\frac{2x-3}{x-1}[/tex] find the asymptotes of this equation.

3. The attempt at a solution

We already know that x=1 is a vertical asymptote from calculating the domain.

the denominator of x-1=0 when x=1.

However I'm not sure what it is they mean when later they go onto say..

[tex]f(x)=\frac{2x-3}{x-1}=\frac{2-\frac{3}{x}}{1-\frac{1}{x}}=\frac{2-0}{1-0}=2 as x\rightarrow \infty[/tex]

What value of x are they using here?

Another more complicated one.

1. The problem statement, all variables and given/known data

I don't know what to do to achieve the same say for this:-

[tex]\frac {5(12x+65)}{x^2-25}}[/tex]

What exactly are they doing when they insert the value of x? It doesn't appear they have the same value for x as they do in the numerator as the denominator? :uhh:

3. The attempt at a solution

This for a graph sketching excercise, by finding certain values you can make a sketch, would I be correct in assuming that there would be a horizontal asymptote in the equation above at y=60, and of course x=-5 and 5 Or am I missing something here?

[tex]\frac{60x+\frac{325}{x}}{x^2-\frac{25}{x^2}}=\frac{60}{1}[/tex]

I think I'm missing something here.

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# Homework Help: Asymptote problem

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