A general question for limits (1 Viewer)

mathsbrain · Jul 23, 2021

Was just trying out a few things with first principles and got stuck at one point:
$\lim_{x\to 0} \cfrac{\sqrt{x^2+x^3}}{x}$
can someone help?

tickboom · Jul 23, 2021

Is it 1? ... My instinct was to try and get that x out of the denominator ...

Trebla · Jul 23, 2021

The limit is undefined (unless x is coming from a specified direction) because the square root of x^2 is |x|, not x

tickboom · Jul 23, 2021

Trebla said:
The limit is undefined (unless x is coming from a specified direction) because the square root of x^2 is |x|, not x

Oooooh yes very good point! ... What about this approach? (I squared top and bottom to simplify instead) ... Or does this approach also have an issue?

Trebla · Jul 23, 2021

I don’t quite follow what you did there, but the answer is definitely not 1. You can plot a graph and see for yourself that the limit is undefined (at least for the two-sided limit).

tickboom · Jul 23, 2021

blyatman said:
You can't just square top and bottom.

It would be like saying:
x = 1
So x^2 = 1
So x = +-1.

Yes good point well presented!

CM_Tutor · Jul 23, 2021

I can similarly rewrite

$\cfrac{\sqrt{x^2 + x^3}}{x} = \cfrac{\sqrt{x^2(1 + x)}}{x} = \cfrac{\sqrt{(-1)^2 x^2(1 + x)}}{x} = \cfrac{\sqrt{(-x)^2(1 + x)}}{x} = \cfrac{-x\sqrt{1 + x}}{x}$

and so get the limit as -1 after cancelling and substituting $x = 0$ .

As Trebla notes,

$\cfrac{\sqrt{x^2 + x^3}}{x} = \cfrac{|x|\sqrt{1 + x}}{x}$

giving the limit as 1 from above and -1 from below, making the limit undefined and the curve $y = \cfrac{\sqrt{x^2 + x^3}}{x}$ discontinuous at $x = 0$ .

A general question for limits (1 Viewer)

mathsbrain

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Trebla

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tickboom

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Trebla

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