Evaluate the following limits or determine that they do not exist. lim(x,y,z)→(2,2,3)x2z−3x2−y2z+3y2xz−3x−yz+3y

Question

Evaluate the following limits or determine that they do not exist.  lim(x,y,z)→(2,2,3)x2z−3x2−y2z+3y2xz−3x−yz+3y

StrycharzT · Accepted Answer

We have to evaluate the limits:  lim(x,y,z)→(2,2,3)x2z−3x2−y2z+3y2xz−3x−yz+3y  Putting x=2, y=2 and z=3 in the function, we get  lim(x,y,z)→(2,2,3)x2z−3x2−y2z+3y2xz−3x−yz+3y=22×3−3×22−22×3+3×222×3−3×2−2×3+3×2  =12−12−12+126−6−6+6  =00  So it is indeterminate form of 00  Solving the limit by factorizing the numerator and denominator,  lim(x,y,z)→(2,2,3)x2z−3x2−y2z+3y2xz−3x−yz+3y=lim(x,y,z)→(2,2,3)x2(z−3)−y2(z−3)x(z−3)−y(z−3)  =lim(x,y,z)→(2,2,3)(z−3)(x2−y2)(z−3)(x−y)  =lim(x,y,z)→(2,2,3)(x−y)(x+y)(x−y)  =lim(x,y,z)→(2,2,3)(x+y)  =2+2  =4  Hence, value of limit is 4.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Evaluate the following limits or determine that they do not exist. lim_{(x,y,z)rightarrow(2,2,3)}frac{x^2z-3x^2-y^2z+3y^2}{xz-3x-yz+3y}

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