2 limit problems

1. Oct 16, 2015

NooDota

1. The problem statement, all variables and given/known data

1. Lim x→0 Sin(x) * sqrt(1 + 1/x^2) Picture: https://i.gyazo.com/2f61c3c09d32447d4190fbdcd3f2f1e5.png

2. Limx→0 Sin(x)/sqrt(x^2 + x^3) Picture: https://i.gyazo.com/b50081d459ed61bcf1d4ae5baecfa7fa.png

2. Relevant equations

3. The attempt at a solution

What I did with the first was turn it into: (sin(x)/x) * (x*sqrt(1 + 1/x^2) which results in (sin(x)/x) * (sqrt(x^2 + 1)

The limit of that is 1 * sqrt(0+1) = 1*1 = 1

Is that correct? Doesn't the lim from right and left give different results?

With the second one, I'm not sure what to do after: Sin(x) * (x^2 + x^3)^-1/2

If I multiply and divide by x I'll get 1 * 0/0 which is undefined, right?

2. Oct 16, 2015

Incand

Yes the limit should be negative when approaching from the left. The thing is when you have $x\sqrt{1+\frac{1}{x^2}} = \sqrt{x^2+1}$ that's only true for $x> 0$.
The second one can be rewritten in two part (if you again take care of the sign). One of them being $\frac{1}{\sqrt{1+x}}$ which has a limit as $x\to 0$.

3. Oct 16, 2015

NooDota

Okay, so what do I do to find the limit from the left?

Also, how do you use the sqrt signs? I didn't find it under the symbols.

Thanks.

Last edited: Oct 16, 2015
4. Oct 16, 2015

Incand

You did it correctly. It's just that you have to remember that for
$x> 0$ we have $x\sqrt{1+\frac{1}{x^2}} = \sqrt{x^2+1}$ and for
$x < 0$ we have $x \sqrt{1+\frac{1}{x^2}} = -\sqrt{x^2+1}$.
You could also write this as $x\sqrt{1+\frac{1}{x^2}} = \text{sign} (x) \sqrt{x^2+1}$ which always is true.
Because the square-root of a number is always positive, that's the definition.
For example when you solve $x^2=1$ you write $x=\pm \sqrt{1}$ since the $\sqrt{1}$ only refers to the positive root. (Sorry If this wasn't what you were confused about, just guessed it may be.)

Do you mean how I write them I text? I use the latex envirement.You can read more about it here https://www.physicsforums.com/threads/introducing-latex-math-typesetting.8997/ . For the sqrt you use that by typing \sqrt{x} inside two # at the start an two # at the end (sorry don't know how to post the code for it, check out the thread i linked!).

5. Oct 16, 2015

Ray Vickson

You can find the limit from the left by writing a correct form that applies when $x < 0$, then take the limit. The form (in (1)) that you wrote (with $\sqrt{x^2+1}$) applies when $x > 0$. Make a change for $x < 0$ and you are almost done. Alternatively, if $x < 0$ you can write $x = -t$, where $t > 0$, then take the limit from the right as $t \to 0$.

Last edited: Oct 16, 2015
6. Oct 16, 2015

NooDota

Okay, thank you both for your help.

I've always dealt with getting x out of the square root, this is the first time I run into a problem where I have to bring it inside instead, hence the confusion.