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adam32885
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 1}\frac{ x1 }{ \sqrt[3]{x+7}2 }\]

adam32885
 2 years ago
Best ResponseYou've already chosen the best response.0I am having trouble with this equation we are given the hint make the substation x+7=t^3

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

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.3Oh now you say you were given that.

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.3\[=\frac{ t^371 }{ t2 }\]

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

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.3And you can get rid of the denominator. Once you do that. Sub back the x's.

adam32885
 2 years ago
Best ResponseYou've already chosen the best response.0so when i make the sub I get \[\lim_{x \rightarrow 1} \frac{ x1 }{ t+2 }\] ?

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.3No you sub the x on the numerator as well.

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

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.3\[t^3=x+7\] make x the subject. \[x=t^37\]

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.3Now sub the x on the numerator with t^37

adam32885
 2 years ago
Best ResponseYou've already chosen the best response.0oh now i get it. pass the duh stamp

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.3Now you can get rid of that denominator.

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

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.3and then you can then sub in x=1 to find the limit.
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