# Limit (indeterminate form)

#### Barioth

##### Member
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
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
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.