Question #160667

Kim(x,y)->(0,0) sinx/y exist.is it true or false. Give reasons for your answer


1
Expert's answer
2021-02-03T15:23:36-0500

Solution:

Lets findtwodifferentpathstoapproachthepointthat\mathrm{Let's \ find\:two\:different\:paths\:to\:approach\:the\:point\:that}

givesdifferentvaluesforthelimit\mathrm {\:gives\:different\:values\:for\:the\:limit}

Case 1: lim(x,y)(0,0)(sin(xy))\lim_{(x,y)\rightarrow (0,0)}(\sin (\dfrac{x}{y})) along x=0x=0 :

=sin(0y)=sin(0)=0=\sin (\dfrac{0}{y})=\sin (0)=0

Case 2: lim(x,y)(0,0)(sin(xy))\lim_{(x,y)\rightarrow (0,0)}(\sin (\dfrac{x}{y})) along x=yx=y :

=sin(yy)=sin(1)=\sin (\dfrac{y}{y})=\sin (1)

Thus, from case 1 and 2, we have that given limit diverges (or does not exist).

Hence, the given statement is false.



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