Is There Another Method to Evaluate This Limit Without Using Taylor Series?

[Barioth]

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

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

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

Thanks for passing by!

 

[I like Serena]

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

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

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 $$\frac y u$$...

 

[Mathwebster]

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 $$\frac y u$$...

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.