Finding the Limit: (1/(x-1)) - (2/(x^2-1))

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Homework Statement


Find the following limit

lim (1/(x-1))-(2/(x^2-1))
x→0


Homework Equations





The Attempt at a Solution


I tried simplifying the equation into one fraction by finding a common denominator. I did this to try to get a denominator where I could plug 1 in without it equaling 0. It didn't work, i ended up with (x^2-2x)/(x^3-x^2-x+1). I feel like I went in the wrong direction from the beginning. If someone could tell me what detail I'm missing it would be greatly appreciated! Thanks!
 
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syd9231 said:

Homework Statement


Find the following limit

lim (1/(x-1))-(2/(x^2-1))
x→0


Homework Equations





The Attempt at a Solution


I tried simplifying the equation into one fraction by finding a common denominator. I did this to try to get a denominator where I could plug 1 in without it equaling 0. It didn't work, i ended up with (x^2-2x)/(x^3-x^2-x+1).
The LCD is x2 -1, not (x - 1) (x2 - 1). All you need to do is multiply the first fraction by x + 1 over itself.
syd9231 said:
I feel like I went in the wrong direction from the beginning. If someone could tell me what detail I'm missing it would be greatly appreciated! Thanks!
 
I did that which left me with (x - 1)/(x^2 - 1) but that still makes the expression undefined when I input 1. I know the answer is 1/2, but I still don't see it. Does the expression mean 1/2 and I'm missing it or is there more computation involved?
 
There's a little bit more to do. Factor the denominator and cancel the common factors in top and bottom.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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