Limit with radicals

[lostmathman]

Hey, I'm trying to find the limit as n goes to infinity of . The answer is 1, but I need to prove it.

If I multiply by the conjugant of the top, I end up with 0, and if I use the bottom I get infinity.. Help please!

 

[DrPhil]

Hey, I'm trying to find the limit as n goes to infinity of View attachment 2632 . The answer is 1, but I need to prove it.

If I multiply by the conjugant of the top, I end up with 0, and if I use the bottom I get infinity.. Help please!

Rationalizing the denominator looks like a good thing to do because it becomes 1. But the numerator becomes

\(\displaystyle (\sqrt{n+2} - \sqrt{n+1})×(\sqrt{n+1} + \sqrt{n}) \)

\(\displaystyle \; \; = \sqrt{(n+2)(n+1)} + \sqrt{(n+2)(n)} - (n + 1) - \sqrt{(n+1)(n)} \)

This still looks like an indeterminant form, \(\displaystyle \infty - \infty \).

Using l'Hospital's rule gave no better results.

How about expanding the square roots using \(\displaystyle \sqrt{1+\delta} \approx \ 1+\delta /2\) ? Begin by dividing every term in the expression by \(\displaystyle \sqrt{n}\):

\(\displaystyle \dfrac{\sqrt{1+2/n} \ - \sqrt{1 + 1/n}}{\sqrt{1 + 1/n} \ - 1}\)

Then apply the square-root approximation. I got the result 1 that way.

 

[MarkFL]

Another approach would be to use the substitution: \(\displaystyle n=\dfrac{1}{u}\) so that you obtain (after multiplying by \(\displaystyle 1=\dfrac{\sqrt{u}}{\sqrt{u}}\)):

\(\displaystyle \displaystyle \lim_{u\to0}\frac{\sqrt{2u+1}-\sqrt{u+1}}{\sqrt{u+1}-1}\)

Now apply L'Hôpital's rule.

 

[lookagain]

.

How about expanding the square roots using \(\displaystyle \sqrt{1+\delta} \approx \ 1+\delta /2\) ? Begin by dividing every term in the expression by \(\displaystyle \sqrt{n}\):

\(\displaystyle \dfrac{\sqrt{1+2/n} \ - \sqrt{1 + 1/n}}{\sqrt{1 + 1/n} \ - 1}\) . . . After this step, then...

L'Hopital's rule can be used at this point. (I did it.)

However, I would have done the substitution that MarkFL made instead,
because with my route, the derivatives and algebra make for
more lines of work and more congested looking steps.

 

[lookagain]

Rationalizing the denominator looks like a good thing to do because it becomes 1. But the numerator becomes

\(\displaystyle (\sqrt{n+2} - \sqrt{n+1})×(\sqrt{n+1} + \sqrt{n}) \)

\(\displaystyle \; \; = \sqrt{(n+2)(n+1)} \ + \ \sqrt{(n+2)(n)} \ - \ (n + 1) \ - \ \sqrt{(n+1)(n)} \)

\(\displaystyle \sqrt{n^2 + 3n + 1 \ } \ + \ \sqrt{n^2 + 2n \ } \ - \ (n + 1) \ - \ \sqrt{n^2 + n \ } \)



You could expand using square roots where \(\displaystyle \sqrt{n^2 + (an + b) \ } \ \approx \ n + \frac{a}{2} :\)**


\(\displaystyle \bigg(n + \dfrac{3}{2}\bigg) \ + \ (n + 1) \ - \ (n - 1) \ - \ \bigg(n + \dfrac{1}{2}\bigg) \ = \) ?








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**In this approximation, all of the terms beyond \(\displaystyle \frac{a}{2} \)
have degree less than or equal to -1.

 

[pka]

Hey, I'm trying to find the limit as n goes to infinity of View attachment 2632 . The answer is 1, but I need to prove it.

If I multiply by the conjugant of the top, I end up with 0, and if I use the bottom I get infinity.. Help please!

\(\displaystyle \dfrac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n+1}-\sqrt{n}}=\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+ \sqrt{n+1}}=\dfrac{\sqrt{1+\frac{1}{n}}+1}{\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}}\)