ysin1/x.... itretive limit and double limit exist or not??
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Yes.
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Anonymous:
are medam ji solution bhi chahiye pura??...how you write yes... explain bhi kr do jra
For x≠0 and y≠0 we have
∣∣∣xsin1y+ysin1x∣∣∣≤∣∣∣xsin1y∣∣∣+∣∣∣ysin1x∣∣∣≤|x|+|y|
So, for
f(x,y)=⎧⎩⎨xsin1y+ysin1x0if x≠0 and y≠0if x=0 or y=0
we have
|f(x,y)|≤|x|+|y|
for all (x,y). Therefore
lim(x,y)→(0,0)f(x,y)=0
by the squeeze theorem.
Be careful that xsin(1/y)≤x is not true in general, but you just need the absolute value and |xsin(1/y)|≤|x| is true (provided y≠0, of course).
∣∣∣xsin1y+ysin1x∣∣∣≤∣∣∣xsin1y∣∣∣+∣∣∣ysin1x∣∣∣≤|x|+|y|
So, for
f(x,y)=⎧⎩⎨xsin1y+ysin1x0if x≠0 and y≠0if x=0 or y=0
we have
|f(x,y)|≤|x|+|y|
for all (x,y). Therefore
lim(x,y)→(0,0)f(x,y)=0
by the squeeze theorem.
Be careful that xsin(1/y)≤x is not true in general, but you just need the absolute value and |xsin(1/y)|≤|x| is true (provided y≠0, of course).
so got ur solution !
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