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Assignment # 1
MTH301 (Spring 2012)
Total marks: 10
Lecture # 1 to 6
Due Date: April 10, 2012
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koi solution send karay ga agr kar skta ha to plz jaldi send kro plzzzzzz
i think we should first put the value x=0 in the function then y=0 and then y=mx.each time we will get different values so the limit of that function do not exist.
i dont think so, becoz we have to "Evaluate the limit along the paths" not prove the continuity...
i agree manan but jab limits exist hi nai kar rai to hum isay evaluate kaisay karain gay?????????
check the solution nd jaldi btana theek ha ya ni
I hate Maths......
Ayesha Kh sharing email address publicly is not allowed..
MTH301 Assignment#1 Solution
See the attached file please
M.Tariq Malik bhai in last step we have 1/ 1+m^2
if we take m=1 then f(x,y)= 1/2 at every point of the line y=x other than (0,0) .
if we take m= -1 then f(x,y)= 1/2 at every point of the line y= -x other than (0,0) .
thus f(x,y) assumes two same values as (x,y) approaches (0,0) along two different paths.
so limit exist... now em confused .. according to dis rule ( bsc book chap# 2, 2.6 exercise) limt exist. or ap ne kha k constant hay is lye limit nai exist krti.. ab ap plzzz mujhy confirm btaye k limt exist krti hy ya nai ta k me apni assignment complt kr sakon