Understanding Limits: Solving for x^2-2|x|/x

  • Thread starter Thread starter ladyrae
  • Start date Start date
  • Tags Tags
    Limit
AI Thread Summary
The limit as x approaches 0 for the expression (x^2 - 2|x|)/x is undefined due to differing left and right limits. As x approaches 0 from the positive side, the limit evaluates to -2, while from the negative side, it evaluates to 2. This discrepancy indicates a discontinuity at x=0, confirming that the limit does not exist. Simplifying the expression to x - 2 is not valid across both sides because |x|/x behaves differently depending on the sign of x. A piecewise approach is necessary to accurately analyze the limit.
ladyrae
Messages
32
Reaction score
0
I’m having trouble understanding this limit

Lim x->0 (x^2-2|x|)/x

I think its undefined but lost marks for that answer on my class assignment.

Also, Why can’t I simplify to x-2?
 
Physics news on Phys.org
The limit exists if and only if the left limit and right limit exist, and are equal to each other.

\lim _{x \rightarrow {0}^{+}} \frac{x^2 - 2|x|}{x}

= \lim _{x \rightarrow {0}^{+}} \frac{x^2 - 2x}{x}

= \lim _{x \rightarrow {0}^{+}} x - 2

= -2


\lim _{x \rightarrow {0}^{-}} \frac{x^2 - 2|x|}{x}

= \lim _{x \rightarrow {0}^{-}} \frac{x^2 + 2x}{x}

= \lim _{x \rightarrow {0}^{-}} x + 2

= 2

Therefore, the limit does not exist.
 
Last edited:
You can also see it graphically by plotting the function. There is a large discontinuity (gap) at x=0.
 
ladyrae said:
I’m having trouble understanding this limit

Lim x->0 (x^2-2|x|)/x

I think its undefined but lost marks for that answer on my class assignment.

Also, Why can’t I simplify to x-2?
Say x = -x. Then
\frac{(-x)^2 - 2|-x|}{-x} = \frac{x^2 - 2x}{-x} = 2 - x​
As you can see, you can't JUST simplify to x - 2. If x > 0, then you can simplify to x - 2.

What happens when x approaches 0 from the positive direction? What happens when x approaches 0 from the negative direction? You should see (if you haven't already) that the limit is undefined.
 
because |x|/x is not equal to 1. It is equal to 1 if x >0 and -1 if x < 0. You can simplify that to x - 2|x|/x. That would help in finding the limit. In order to find the limit you will have to break that up into a piecewise function and than look at the limit as it approaches 0 from the left and the right. And ill give you a hint, there is no limit as x --> 0.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top