# Thread: Limits at infinity?

1. ## 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?

2. ## Re: Limits at infinity?

For the first one:
$\lim_{x\to \infty} \frac{\sin(2x)}{x}$

If $x \to \infty$, then $\sin(2x)$ goes to? ... (think of the range)

3. ## Re: Limits at infinity?

It seems for my first question the limit would be zero.

4. ## Re: Limits at infinity?

Originally Posted by homeylova223
2. As x approaches infinity
((1))/((2x+sin(x))
If $x\ge 3$ then $0<2x-1\le 2x+\sin(x)\le 2x+1.$

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

If $x\to \infty$ what happens to $\frac{1}{2x+\sin(x)} ~?$

5. ## Re: Limits at infinity?

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 $x \to +\infty$ and $x \to -\infty$.

6. ## 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.

7. ## Re: Limits at infinity?

Can you go further with the hint I gave?
$\lim_{x\to + \infty} \left(x+\sqrt{x^2+3x}\right)=+\infty+\infty=\infty$
For $x \to - \infty$ you'll get the indetermined form $\infty-\infty$ so multiply and divide the limit by $x-\sqrt{x^2+3}$ ...

8. ## 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.

9. ## Re: Limits at infinity?

What do you think of:
$\lim_{x\to -\infty} \frac{(x+\sqrt{x^2+3})\cdot (x-\sqrt{x^2+3})}{x-\sqrt{x^2+3}}=\lim_{x\to -\infty}\frac{x^2-x^2-3}{x-\sqrt{x^2+3}}=\lim_{x\to -\infty}\frac{-3}{-\infty-\infty}=\lim_{x\to -\infty} \frac{-3}{-(\infty+\infty)}=0$

Do you understand?

10. ## 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

11. ## Re: Limits at infinity?

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:

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

So now you need only worry about:

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

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

CB