Check out some similar questions!

pop000 · Jan 6, 2011, 09:39 AM

be the Function (in the picture) a,b are Positive, is known that y=1 is a Horizontal Asymptote OF the Function :limf(x)=9 when x-->o+.

I got this answers and I need to know what is the sum of a+b:

1)3
2)4
3)5
4)6
5)7
thanks.

ebaines · Jan 6, 2011, 10:13 AM

I think you have an error here. Given the fiunction:

$<br /> f(x) = \frac {x \sin x+ (ax+b)^2}{x^2+1}<br />$

The limit as x approaches infinity is:

$<br /> \displaystyle \lim_{x \to \infty} f(x)= \displaystyle \lim_{x \to \infty} \frac {x \sin x } {x^2+1} \ + \displaystyle \lim_{x \to \infty} \frac {(ax+b)^2} {x^2+1} = \ 0 \ + \displaystyle \lim_{x \to \infty} \frac {a^2x^2 + 2 abx + b^2} {x^2+1} = a^2<br />$

So if the limit approaches y = 1, then you know that $a^2 = 1$ It does not matter what the value of b is.

**EDIT** Unfortunately I solved the wrong problem! Sorry 'bout that.

galactus · Jan 6, 2011, 10:15 AM

Since we are told there is a horizontal asymptote, then we must have:

$\lim_{x\to \infty}\frac{xln(x)+(ax+b)^{2}}{x^{2}+1}$

If we evaluate this limit it is

$a^{2}$

But we were told this limit is 1, so

$a^{2}=1$

$\fbox{a=1}$

Expanding f(x) gives us a term which is:

$\frac{b^{2}}{x^{2}+1}$ (all other terms tend to 0 as x approaches 0).

Since x=0 and this term must equal 9, then:

$b^{2}=9\Rightarrow \fbox{b=3}$

The value of b determines how high up the y-axis the graph goes.

For instance, if b=4, then $\lim_{x\to 0^{+}}f(x)=16$

Thus, the sum is 4.

ebaines · Jan 6, 2011, 10:26 AM

Ahh , I misread the post! I thought it was (a) the limit going to infinity (not 0) and (b) I read "xln(x) as "sin(x)." It's neen a long day...

galactus · Jan 7, 2011, 05:50 AM

I read "xln(x) as "sin(x)."

You know, I done that as well. :confused:

Then, I looked again and saw the xlnx.

pop000 · Jan 7, 2011, 05:54 AM

Thanks to both of you and yes we all mistake sometime :-)

Still thank you People.