Answer:yes
Step-by-step explanation: 73x 10 -73 essentially
Answer:
This approach to (0,0) also gives the value 0
Step-by-step explanation:
Probably, you are trying to decide whether this limit exists or not. If you approach through the parabola y=x², you get
![\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{\sqrt{x^2+y^2}}=\lim_{(x,x^2)\rightarrow (0,0)}\frac{xx^2}{\sqrt{x^2+(x^2)^2}}=\lim_{x\rightarrow 0}\frac{x^3}{|x|\sqrt{1+x^2}}=0](https://tex.z-dn.net/?f=%5Clim_%7B%28x%2Cy%29%5Crightarrow%20%280%2C0%29%7D%5Cfrac%7Bxy%7D%7B%5Csqrt%7Bx%5E2%2By%5E2%7D%7D%3D%5Clim_%7B%28x%2Cx%5E2%29%5Crightarrow%20%280%2C0%29%7D%5Cfrac%7Bxx%5E2%7D%7B%5Csqrt%7Bx%5E2%2B%28x%5E2%29%5E2%7D%7D%3D%5Clim_%7Bx%5Crightarrow%200%7D%5Cfrac%7Bx%5E3%7D%7B%7Cx%7C%5Csqrt%7B1%2Bx%5E2%7D%7D%3D0)
It does not matter if x>0 or x<0, the |x| on the denominator will cancel out with an x on the numerator, and you will get the term x²/(√(1+x²) which tends to 0.
If you want to prove that the limit doesn't exist, you have to approach through another curve and get a value different from zero.
However, in this case, the limit exists and its equal to zero. One way of doing this is to change to polar coordinates and doing a calculation similar to this one. Polar coordinates x=rcosФ, y=rsinФ work because the limit will only depend on r, no matter the approach curve.
Answer:
Never is the correct answer.
Step-by-step explanation:
The 1st and 4th terms are the extremes, the 2nd and 3rd terms are the means.
Answer:
The Answer is 18 then 25 then 21 then 20
Step-by-step explanation: