Understanding Limits: Why |h|/h Approaches -1 as h Approaches 0

  • Thread starter Thread starter ratios
  • Start date Start date
  • Tags Tags
    Limit
ratios
Messages
3
Reaction score
0
Hello all,

I don't understand why |h|/h will become -1 as h approches 0 from the left. I can see how the numerator becomes negative, as |x| = -x if x< 0, but shouldn't the denominator become negative also since h < 0 when we approach 0 from the left? So why isn't the answer -h/-h = 1?
 
Physics news on Phys.org
The numerator is always positive. It's true that |x|=-x when x<0, but in that case -x>0, and |x|/x=-x/x=-1.
 
ratios said:
I can see how the numerator becomes negative, as |x| = -x if x< 0
You see wrong. Here, -x is positive.
 
Okay I see. Thanks for the help!
 
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...
Back
Top