Question

lim (x,y,z) ->(0,0,0) [e^sqrt(x^ 2 +y^ 2 +z^ 2) -cos(x^ 2 +y^ 2 +z^ 2 ) ]÷sqrt( x^ 2 +y^ 2 +z^ 2)​

Answer 1

Answer:

L.H.S. =sin

2

x+cos

2

y≤2          [∵sin

2

x≤ and cos

2

y≤1]

R.H.S. =2sec

2

z≥2

Hence, L.H.S.=R.H.S. only when sin

2

x=1, cos

2

y=1 and 2sec

2

z=2. Thus,

   

cos

2

x=0, sin

2

y=0, cos

2

z=1

⇒   cosx=0, siny=0, sinz=0

   

x=(2m+1)

2

π

​

, y=nπ and z=tπ, where m, n and t are integers

Step-by-step explanation:

Answer 2

Step-by-step explanation:

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