# Limits at infinity?

• Jul 31st 2011, 01:33 PM
homeylova223
Limits at infinity?
Can anyone help me with these limits

1. As x approaches infinity

((sin2x))/((x))

2. As x approaches infinity

((1))/((2x+sin(x))

3. As x approaches infinity

((x+ square root(x^2+3)) I have some work for this one

((x+ square root(x^2+3))/((1)) I rationalized the function

((x^2+x^2+3))/((x+square root(x^2+3))

Now my main issue is how would I get a common denominator what should I chose?(Crying)
• Jul 31st 2011, 01:52 PM
Siron
Re: Limits at infinity?
For the first one:
$\displaystyle \lim_{x\to \infty} \frac{\sin(2x)}{x}$

If $\displaystyle x \to \infty$, then $\displaystyle \sin(2x)$ goes to? ... (think of the range)
• Jul 31st 2011, 02:18 PM
homeylova223
Re: Limits at infinity?
It seems for my first question the limit would be zero.
• Jul 31st 2011, 02:53 PM
Plato
Re: Limits at infinity?
Quote:

Originally Posted by homeylova223
2. As x approaches infinity
((1))/((2x+sin(x))

If $\displaystyle x\ge 3$ then $\displaystyle 0<2x-1\le 2x+\sin(x)\le 2x+1.$

That gives $\displaystyle \frac{1}{2x+1}\le \frac{1}{2x+\sin(x)} \le \frac{1}{2x-1}$

If $\displaystyle x\to \infty$ what happens to $\displaystyle \frac{1}{2x+\sin(x)} ~?$
• Jul 31st 2011, 04:15 PM
Siron
Re: Limits at infinity?
Quote:

Originally Posted by homeylova223
It seems for my first question the limit would be zero.

Correct!
For the last one, make a difference by calculating $\displaystyle x \to +\infty$ and $\displaystyle x \to -\infty$.
• Jul 31st 2011, 04:33 PM
homeylova223
Re: Limits at infinity?
Seeing Plato inequality the limit would have to approach x at zero.

Now all I have left is my third question.
• Jul 31st 2011, 04:38 PM
Siron
Re: Limits at infinity?
Can you go further with the hint I gave?
$\displaystyle \lim_{x\to + \infty} \left(x+\sqrt{x^2+3x}\right)=+\infty+\infty=\infty$
For $\displaystyle x \to - \infty$ you'll get the indetermined form $\displaystyle \infty-\infty$ so multiply and divide the limit by $\displaystyle x-\sqrt{x^2+3}$ ...
• Jul 31st 2011, 07:18 PM
homeylova223
Re: Limits at infinity?
Hmm I am slight confused

if I multiply by ((x+square root(x^2+3))

I get ((x^2+x^2+3))/((x+square root(x^2+3)) I used the denominator x^2 for the numerator

((x^2+x^2+3))/((x^2)) to get ((1+1+3/x^2))

But what could I use for the denominator I tried x but I think it incorrect.
• Jul 31st 2011, 07:40 PM
Siron
Re: Limits at infinity?
What do you think of:
$\displaystyle \lim_{x\to -\infty} \frac{(x+\sqrt{x^2+3})\cdot (x-\sqrt{x^2+3})}{x-\sqrt{x^2+3}}=\lim_{x\to -\infty}\frac{x^2-x^2-3}{x-\sqrt{x^2+3}}=\lim_{x\to -\infty}\frac{-3}{-\infty-\infty}=\lim_{x\to -\infty} \frac{-3}{-(\infty+\infty)}=0$

Do you understand?
• Jul 31st 2011, 07:49 PM
homeylova223
Re: Limits at infinity?
I am wondering when you get ((x^2-x^2-3))/((x-square root(x^2+3))

How do you get -infinity-infinity
• Jul 31st 2011, 08:13 PM
CaptainBlack
Re: Limits at infinity?
Quote:

Originally Posted by homeylova223
I am wondering when you get ((x^2-x^2-3))/((x-square root(x^2+3))

How do you get -infinity-infinity

You don't, that is an invalid move.

You have:

$\displaystyle \lim_{x \to \infty}\frac{-3}{x-\sqrt{x^2+3}}=\lim_{x \to \infty}\frac{-3}{x-x\sqrt{1+3/x^2}}$

So now you need only worry about:

$\displaystyle \lim_{x \to \infty}\left(x-x\sqrt{1+3/x^2}\right)$

and this can be determined using a power series expansion of the square root.

CB