Limit question limit (x->0) 1/|x|

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limit (x->0) 1/|x|
 
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the limit doesn't exist.
lim (x-> o+) = + ∞
lim (x-> o-) = - ∞

so limit of the function at the point zero is not existent by definition.
 
AdrianZ said:
the limit doesn't exist.
lim (x-> o+) = + ∞
lim (x-> o-) = - ∞

so limit of the function at the point zero is not existent by definition.

Not true. The function is 1/|x| not 1/x
 
Mentallic said:
Not true. The function is 1/|x| not 1/x

aah yea you're right. didn't notice the abs function there. anyway, when x approaches zero from left (and right) the function approaches +∞. so the limit is +∞.

that means for all positive deltas and M's if |x - 0| < delta, then f(x) > M. for any large M you like. intuitively you can say: the larger you want M to be, the smaller you need to take delta.
 
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Which still means that there is no limit. Saying "\lim f(x)= \infty" is just a way of saying that the limit does not exist for a particular reason.
 
well, I think if we're working on extended real number line then it doesn't hurt to say the limit of the function is +∞ because +∞ ∈ R*. and plus the epsilon, delta definition works fine with this example. correct me if I'm wrong.
 
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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