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adam32885Best ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow 1}\frac{ x1 }{ \sqrt[3]{x+7}2 }\]
 one year ago

adam32885Best ResponseYou've already chosen the best response.0
I am having trouble with this equation we are given the hint make the substation x+7=t^3
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
Rationalise the denominator by multiplying by \[\frac{ \sqrt[3]{(x+7)^2}+2 }{ \sqrt[3]{(x+7)^2}+2 } \]
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
Oh now you say you were given that.
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
\[=\frac{ t^371 }{ t2 }\]
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
x+7=t^3 as x > 1 t > 2
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
And you can get rid of the denominator. Once you do that. Sub back the x's.
 one year ago

adam32885Best ResponseYou've already chosen the best response.0
so when i make the sub I get \[\lim_{x \rightarrow 1} \frac{ x1 }{ t+2 }\] ?
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
No you sub the x on the numerator as well.
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
a^3  b^3 = (ab) (a^2 +ab + b^2
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
\[t^3=x+7\] make x the subject. \[x=t^37\]
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
Now sub the x on the numerator with t^37
 one year ago

adam32885Best ResponseYou've already chosen the best response.0
oh now i get it. pass the duh stamp
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
Now you can get rid of that denominator.
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
Once you do that, remember to replace the t's with \[\sqrt[3]{x+7}\]
 one year ago

AzteckBest ResponseYou've already chosen the best response.3
and then you can then sub in x=1 to find the limit.
 one year ago
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