How to Solve Limit Problems with Square Roots and Logarithms?

[danni7070]

I was just wondering if this was the right way to solve this limit problem.

$$\lim_{x\rightarrow\infty} (\sqrt{x+1} - \sqrt{x})^\frac{1}{ln(x)}$$

Multiply both sides...

$$(\frac{1}{\sqrt{x+1}+\sqrt{x}})^\frac{1}{ln(x)} = 0^0$$

which is undefined.

Any suggestions?

 

[mjsd]

the correct way to do this is to take the natural log of both sides:
L = lim ...
Log L = Log (lim...)

etc

 

[HallsofIvy]

I was just wondering if this was the right way to solve this limit problem.

$$\lim_{x\rightarrow\infty} (\sqrt{x+1} - \sqrt{x})^\frac{1}{ln(x)}$$

Multiply both sides...

$$(\frac{1}{\sqrt{x+1}+\sqrt{x}})^\frac{1}{ln(x)} = 0^0$$

which is undefined.

Any suggestions?

Multiply both sides of WHAT? by WHAT?

As mjs said, you can simplify this by taking the logarithm. Of course that means you get the logarithm of the answer you want.

 

[dynamicsolo]

I believe the OP meant the numerator and denominator of the expression in parentheses was multiplied by the "conjugate factor" ("both sides" of the ratio line...). That's a reasonable first step before going over to the logarithm, since you're going to have to face that indeterminate difference soon enough anyway.

I'll point out that after you perform a "L'Hopital" on the resulting indeterminate product converted to a ratio, you will still have an indeterminate ratio, which you will not want to use L'Hopital's Rule on again. Instead, a simpler approach will finish the job (and the answer is finite... don't forget to undo the logarithm at the end...).

 

[danni7070]

Yeah, thanks dynamicsolo that's exactly what I was doing. I always forget what's the denominator and numerator cause I've learned it my own language and it's like there isn't any room for more names :)

I'm on a pause in mathematics. The test is over and it was ok. But for now I have to concentrate on other things.

Thank you all dynamic, halls, Dick and others.