how to finf the horizontal asymptot of
f(x) = x^2-1/x^2+x-2
Hello, wutever
The traditional approach . . .Find the horizontal asymptote of: .$\displaystyle f(x) \:= \:\frac{x^2-1}{x^2+x-2}$
Find: .$\displaystyle \lim_{x\to\infty} f(x)$
If the limit has a finite value $\displaystyle a$, the horizontal asymptote is: .$\displaystyle y = a$
We have: .$\displaystyle \lim_{x\to\infty}\,\frac{x^2-1}{x^2+x-2}$
Divide top and bottom by the highest power of $\displaystyle x$ in the denominator:
. . $\displaystyle \lim_{x\to\infty}\,\frac{\frac{x^2}{x^2} - \frac{1}{x^2}}{\frac{x^2}{x^2} + \frac{x}{x^2} - \frac{2}{x^2}} \;=\;\lim_{x\to\infty}\,\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x} - \frac{1}{x^2}} \;=\;\frac{1 - 0}{1 + 0 - 0} \;=\;1$
Therefore, the horizontal asymptote is: .$\displaystyle y \,=\,1$
thx for the reply
i have another one (sqr root of 9x^2+5x-2) / 2x^3+1
how to find its horizontal asymptot..
i have the answer which is (sqr root of 9) / 2
so is it a rule that the asymptot is the cooeficent of x in the denominator over the cooficent of x in the numenerator ???
clarify, is this $\displaystyle f(x) = \frac {\sqrt{9x^2 + 5x - 2}}{2x^3 + 1}$ ?
if that's the case, your answer is wrong
exactly the same way, find $\displaystyle \lim_{x \to \infty}f(x)$ and $\displaystyle \lim_{x \to - \infty} f(x)$how to find its horizontal asymptot..
no, only when the highest power in the numerator is the SAME as the highest power in the denominator, then the answer is the ratio of the two coefficients, otherwise, not necessarilyi have the answer which is (sqr root of 9) / 2
so is it a rule that the asymptot is the cooeficent of x in the denominator over the cooficent of x in the numenerator ???
in that case, the answer would be that you have 2 horizontal asymptotes: $\displaystyle \pm \frac 32$
we want $\displaystyle \lim_{x \to \infty}f(x)$ and $\displaystyle \lim_{x \to - \infty}f(x)$ where $\displaystyle f(x) = \frac {\sqrt{9x^6 + 5x - 2}}{2x^3 + 1}$
now as x goes to infinity, only the highest powers matter, since doing trivial things like subtracting 2 or adding 1 would be inconsequential "at" infinity. and so would adding 5x when it is contending with 9x^6.
thus: $\displaystyle \lim_{x \to \infty} \frac {\sqrt{9x^6 + 5x - 2}}{2x^3 + 1} = \lim_{x \to \infty} \frac {\sqrt{9x^6}}{2x^3}$
$\displaystyle = \lim_{x \to \infty} \frac {\sqrt{9x^4x^2}}{2x^3}$
$\displaystyle = \lim_{x \to \infty} \frac {3x^2 \sqrt{x^2}}{2x^3}$
$\displaystyle = \lim_{x \to \infty} \frac {3|x|}{2x}$ ........since $\displaystyle |x| = \sqrt{x^2}$
$\displaystyle = \lim_{x \to \infty} \frac 32 \frac {|x|}{x}$
so we have: $\displaystyle \frac 32 \frac {|x|}x \to \left \{ \begin{array}{lr} ~~\frac 32 & \mbox{ if } x \ge 0 \mbox{ that is, } x \to \infty \\ & \\ - \frac 32 & \mbox{ if } x < 0 \mbox{ that is, } x \to - \infty \end{array} \right.$