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last discontinuity question... :)
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\[f(x) = \frac{ 5 }{ \log_3 x }, x>0\]
i know that if i plugin x = 3, the log base 3 and the number 3 would cancel out...leaving us with just...f(3) = 5
it is continuous
Set log3(x) to 0. Use the base changing formula. that would be logx/log5 = 0 then logx=0 x=1 so it's discontinuous at 1. also log can never have negative. so it is continuous x>0 and x≠1
omg good thinking ...i didn't try to think that way... good catch bro :)
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Kya ya aakhire savaal hai?
iss exercise/mushk ka akhri swal tha lekin mere pass aur bohot sare sawalaat hein :)
theek
mein zara likh loon bhai...aik menut
haan theek
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