Evaluating the limit of a function

moouers

1. Homework Statement
I'm not sure how to type "lim x-> 0 out using this forum's math symbols...

lim x→0 [$\frac{1}{x\sqrt{1+x}}$-$\frac{1}{x}$]

2. Homework Equations

3. The Attempt at a Solution

Honestly, I'm just not seeing how to manipulate the equation to get it to a point where I can find the solution. This isn't graded and isn't even for any class I'm in...I'm just curious. It's been 10 years since I've done anything like this, so I'm trying to get up to speed before I take Calculus again in the fall.

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genericusrnme

Since x is going to zero, I'd try a taylor series about x=0 ;)

moouers

Haha, well, I'd love to. But since this is in one of the first few chapters in the book, long before Taylor series is introduced, there must be another way and I'd like do it that way first.

genericusrnme

oh.. :p

In that case, try getting everything into one term, with a common denominator and see what happens

moouers

I tried that and was just as lost. Maybe I did it wrong, but I got:

lim x→0 [$\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}$]

If I did that correctly, I still don't know where to go from there.

genericusrnme

also try multiplying by 1 in a nice way :3

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moouers

Would you mind showing my the first step to take, so I can go on from there? I would appreciate it.

Dick

Homework Helper
I tried that and was just as lost. Maybe I did it wrong, but I got:

lim x→0 [$\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}$]

If I did that correctly, I still don't know where to go from there.
Now try multiplying numerator and denominator by $1+\sqrt{1+x}$. Expand out the numerator.

alanlu

No, the expansion was done correctly.

Try rationalizing the numerator by multiplying by $\frac{1 + \sqrt{1 + x}}{1 + \sqrt{1 + x}}$

Also, Taylor series are used for approximating functions, not evaluating limits.

Edit: ninjas on this board!

moouers

Wow, so that was the trick. I carried out the math from there and caclulator/book confirms the answer. Thank you all so much. That was far simpler than it seemed.

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