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jpsmarinho
 4 years ago
lim sqrt(x^29)/(2x6) when x goes to infinity. I use a software and discover that the lim equals to 1/2, but, when I do, II only encounter the result 1/2
jpsmarinho
 4 years ago
lim sqrt(x^29)/(2x6) when x goes to infinity. I use a software and discover that the lim equals to 1/2, but, when I do, II only encounter the result 1/2

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jpsmarinho
 4 years ago
Best ResponseYou've already chosen the best response.0Ah but has square root in the x^2 9

jpsmarinho
 4 years ago
Best ResponseYou've already chosen the best response.0The lim is the following: {(x^29)^(1/2)}/(2x6)

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0well it should be 1/2 only cant be 1/2 as for x tending to (a very large no.) ,,the expression gets negative,,clearly.. you can justify this as you may 4 = (2)^2 and send 4 under sqrt sign with  aside ..

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0if you understand what am saying.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For positive infinity it is 1/2 for negative infinity it is 1/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jpsmarinho This is answer of your question..

jpsmarinho
 4 years ago
Best ResponseYou've already chosen the best response.0What you did in first step? I don't understand.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1338756191105:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jpsmarinho Now remove the square y and get the radical over whole function .

jpsmarinho
 4 years ago
Best ResponseYou've already chosen the best response.0Other doubt: why I can not consider power the numerator and denominator without equal the expression to some variable?
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