Finding limits with absolute values

Alain12345
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I need help finding limits. I know it's pretty simple most of the time... I know that for example if the lim x--> 3 of x-3, you just plug 3 for x... what do I do if it's the absolute value of x-3?

I know you guys like to see that I've tried to solve the problem, but there's not much I can show you in this case.

Thanks.
 
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you have to look at the right and left hand limits (i.e. lim as x approaches a from the left, etc...)
 
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If you have an expression that is in indeterminate form, you may try to factor it out. If you cannot factor it, then try multiply it with its conjugate pair.
 
For absolute value problems, you can write it as a piecewise defined function. i.e. instead of y = abs(x), you can write y = { x, x>=0; -x,x<0

Then, as courtrigard said, look at the left hand and right hand limits.
 
Finding d limit of the absolute value of x-3, divided by x-3, for x approuches 0.

lim [x-3]/(x-3)
x->0
Principle [x-3]= {(x-3) if x>4}
{-(x-3) if x<4}

Remember that limit olny exist only if the limit from the negetive side is equal to the limit from the positive side as x approuches the given value!

= lim (x-3)/(x-3) = 1 and =lim -(x-3)/(x-3) = -1
x=>0+ x=>0-
we can conclude by saying
1 is not equal to -1
The limit of [x-3]/x-3 as x=>0 does not exist!
 
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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