I got problem with nr 39. If i understand right there shall be at limit x->2 because denominator. With other words if i understand correct i shall solve limit x->2 (2x+1)/(x-2) well i basicly got problem to solve that limit.

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- Feb 12th 2013, 02:02 AMPetrusHorizontal and vertical asymptotes
I got problem with nr 39. If i understand right there shall be at limit x->2 because denominator. With other words if i understand correct i shall solve limit x->2 (2x+1)/(x-2) well i basicly got problem to solve that limit.

- Feb 12th 2013, 02:18 AMearbothRe: Horizontal and vertical asymptotes
1. A vertical asymptote occurs if the denominator of a quotient equals zero and the numerator unequals zero. With your example:

$\displaystyle y = \frac{2x+1}{x-1}$

you'll get a vertical asymptote at x = 1

2. Re-write the equation by synthetic division:

$\displaystyle y = \frac{2x+1}{x-1}~\implies~ \boxed{y = 2 + \frac3{x-1}}$

Since $\displaystyle \lim_{x \to \infty}\left(\frac3{x-1} \right) = 0$

you'll get

$\displaystyle \lim_{x \to \infty} y = 2$

3. The line with the equation y = 2 is the horizontal asymptote. - Feb 12th 2013, 02:28 AMPetrusRe: Horizontal and vertical asymptotes
Thanks!

Just wanna correct u it was x-2 on denominator so vertical asymptot at x=2:)

so we rewrite the equation as 2+5/(x-2)

Edit: just a question. Should i also check lin x->-infinity aswell? In this problem it Will still be 2 but i should also check when lim x->-infinity? I am correct? - Feb 12th 2013, 03:07 AMPetrusRe: Horizontal and vertical asymptotes
On nr 43. I get horizontal asymptotes at infinity and -infinity. Shall i write that?? I got vertical at x=5

- Feb 12th 2013, 05:28 AMearbothRe: Horizontal and vertical asymptotes
1. Factorize the denominator (and in this case the numerator too):

$\displaystyle y = \frac{x^3-x}{x^2-6x+5} = \frac{x(x^2-1)}{(x-1)(x-5)}= \frac{x(x-1)(x+1)}{(x-1)(x-5)}$

That means you'll get a vertical asymptote at x = 5 and a hole in the graph at x = 1.

2. Re-write the term of the function by synthetic division:

$\displaystyle y = \frac{x^3-x}{x^2-6x+5} ~\implies~y = \underbrace{x+5}_{\text{term of the asymptote}}+\frac{25x-25}{x^2-6x+5}$

That means your function has a slanted asymptote $\displaystyle y = x+5$ - Feb 12th 2013, 05:52 AMPetrusRe: Horizontal and vertical asymptotes
- Feb 12th 2013, 09:57 AMMINOANMANRe: Horizontal and vertical asymptotes
IF limf(x)/x =a when x goes to infinity or minus infinity and lim(f(x)-ax)=b when x goes to infinity or - infinity then the line y=ax+b is a slant asymptote of f(x).

MINOAS - Feb 12th 2013, 11:03 AMearbothRe: Horizontal and vertical asymptotes
- Feb 12th 2013, 11:28 AMPetrusRe: Horizontal and vertical asymptotes
- Feb 12th 2013, 11:44 AMMINOANMANRe: Horizontal and vertical asymptotes
Some functions not all ...have asymptotes that are not vertical or horizontal .these are slant asymptotes...depending on your math carricculum you may or may not learnt them...the above theorem I mentioned is used to track them.

Minoas