Given #(u^4 + 3u + 6)^(1/2)# how do you find the limit as u approaches -2?

1 Answer
Truong-Son N.
Jun 1, 2016

Just plug it in.

#color(blue)(lim_(u->-2) (u^4 + 3u + 6)^"1/2")#

#= sqrt((-2)^4 + 3(-2) + 6)#

#= sqrt(16)#

#= color(blue)(4)#

All this means is you travel along the #u# axis and approach the value #u = -2# from either side (left/right). The answer tells you that the #v# value is #4#.

In other words, the coordinate #(u,v) = (-2, 4)# exists in #v(u) = (u^4 + 3u + 6)^"1/2"#.

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