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Limit (indeterminate form)

Barioth

Member
Jan 17, 2013
52
Hi, I have a question that is very close the the one of the OP so I tough I should post in here instead of making a new thread. (Hope no one mind )

Let's say

\(\displaystyle \lim_{x->\pm\infty}x(log \sqrt{x} - log(\sqrt{x}-y)-\frac{y}{\sqrt{x}} )=\frac{y^2}{2}\)

Now I could use taylor series to evaluate it, is there another way? (I find using taylor series very long)

Thanks for passing by!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hi, I have a question that is very close the the one of the OP so I tough I should post in here instead of making a new thread. (Hope no one mind )

Let's say

\(\displaystyle \lim_{x->\pm\infty}x(log \sqrt{x} - log(\sqrt{x}-y)-\frac{y}{\sqrt{x}} )=\frac{y^2}{2}\)

Now I could use taylor series to evaluate it, is there another way? (I find using taylor series very long)

Thanks for passing by!
Well, let's substitute $u = \sqrt x$.
Note that the limit can only exist if x is positive.

Then you get
$$\lim_{u \to \infty}u^2\left(\log u - \log(u-y)-\frac{y}{u} \right)
= \lim_{u \to \infty}u^2\left(\log u - \log\big( u(1-\frac y u)\big)-\frac{y}{u} \right)
= \lim_{u \to \infty}u^2\left(- \log\big(1-\frac y u\big)-\frac{y}{u} \right)
$$
Now do a Taylor expansion with \(\displaystyle \frac y u\)...
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Well, let's substitute $u = \sqrt x$.
Note that the limit can only exist if x is positive.

Then you get
$$\lim_{u \to \infty}u^2\left(\log u - \log(u-y)-\frac{y}{u} \right)
= \lim_{u \to \infty}u^2\left(\log u - \log\big( u(1-\frac y u)\big)-\frac{y}{u} \right)
= \lim_{u \to \infty}u^2\left(- \log\big(1-\frac y u\big)-\frac{y}{u} \right)
$$
Now do a Taylor expansion with \(\displaystyle \frac y u\)...
You could at this point let $1 - \dfrac{y}{u} = e^p$ so that $u \to \infty$ gives $p \to 0$ and turn the limit into one that you can use L'Hopital's rule on.