"Radical" Limit Problem

E

eric_f

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Apr 23, 2013
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33
Hi all,

Here's a scan of where I'm spinning my wheels on this one. I've been playing with it for a good two hours and am stuck past the axle. No matter how I manipulate it, it always seems to be an indeterminate form...

scan0002.jpg

Any tips to get me rolling along? I know it's probably something simple I'm overlooking. :)

Thanks!
 
eric_f said:
Hi all,

Here's a scan of where I'm spinning my wheels on this one. I've been playing with it for a good two hours and am stuck past the axle. No matter how I manipulate it, it always seems to be an indeterminate form...

View attachment 2832

Any tips to get me rolling along? I know it's probably something simple I'm overlooking. :)

Thanks!
Click to expand...
Your work is correct - it definitely IS an indeterminate form, 0/0. Do you know l'Hospital's rule? If so, now is the time to use it. If not, ...

BTW, you could have saved a step or two by noticing a common factor sqrt(x) in the two denominators, so the LOWEST common denominator is sqrt(x)×sqrt(x+1).
 
DrPhil said:
Your work is correct - it definitely IS an indeterminate form, 0/0. Do you know l'Hospital's rule? If so, now is the time to use it. If not, ...

BTW, you could have saved a step or two by noticing a common factor sqrt(x) in the two denominators, so the LOWEST common denominator is sqrt(x)×sqrt(x+1).
Click to expand...


No, I don't know l'Hospital's rule yet... I'm only two weeks into Calculus 1. This was presented as a challenge problem, but I think I've got it...

scan0004.jpg

I'm sure I added more steps than necessary, but everything that went through my brain went on paper. :mrgreen:
 
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lim⁡n→∞\displaystyle \lim_{n\rightarrow \infty }(nnnn+1\displaystyle \frac{n}{\sqrt{n}}\frac{n}{\sqrt{n+1}})

where n=1x\displaystyle n=\frac{1}{x}

still indeterminate :(
 
Last edited:
Hello, eric_f!

DrPhil gave you excellent advice.

lim⁡x→0+(1x−1x2+x)\displaystyle \displaystyle\lim_{x\to0^+}\left(\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2+x}}\right)
Click to expand...


We have: .1x−1x(x+1)  =  1x−1xx+1\displaystyle \dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x(x+1)}} \;=\;\dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x}\sqrt{x+1}}

. . . . . . =  1x⋅x+1x+1−1xx+1  =  x+1−1xx+1\displaystyle \displaystyle=\;\frac{1}{\sqrt{x}}\cdot\frac{\sqrt{x+1}}{ \sqrt{x+1}} - \frac{1}{\sqrt{x}\sqrt{x+1}} \;=\;\frac{\sqrt{x+1} - 1}{\sqrt{x}\sqrt{x+1}}


Multiply by x+1+1x+1+1\displaystyle \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}

. . x+1−1xx+1⋅x+1+1x+1+1  =  (x+1)−1xx+1(x+1+1)\displaystyle \displaystyle\frac{\sqrt{x+1}-1}{\sqrt{x}\sqrt{x+1}}\cdot\frac{ \sqrt{x+1}+1}{ \sqrt{x+1} + 1} \;=\; \frac{(x+1)-1}{\sqrt{x}\sqrt{x+1}(\sqrt{x+1}+1)}

. . . . . =  xxx+1(x+1+1)  =  xx+1(x+1+1)\displaystyle \displaystyle=\;\frac{x}{\sqrt{x}\sqrt{x+1}(\sqrt{x+1}+1)} \;=\; \frac{\sqrt{x}}{\sqrt{x+1}(\sqrt{x+1} + 1)}


. . lim⁡x→0+xx+1(x+1+1)  =  00+1(0+1+1)  =  02  =  0\displaystyle \displaystyle \lim_{x\to0^+} \frac{\sqrt{x}}{\sqrt{x+1}(\sqrt{x+1} + 1)} \;=\;\frac{\sqrt{0}}{\sqrt{0+1}(\sqrt{0+1} + 1)} \;=\;\frac{0}{2} \;=\;0
 
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