SOLUTION: how do you prove that the limit of x approaching 18 of (36-2x)/((sqrt2x)-6)= -12

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Question 951039: how do you prove that the limit of x approaching 18 of (36-2x)/((sqrt2x)-6)= -12
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Use a substitution, let u=sqrt%282x%29
Then,
u%5E2=2x
So then,
%2836-u%5E2%29%2F%28u-6%29=%28%286-u%29%286%2Bu%29%29%2F%28u-6%29
%2836-u%5E2%29%2F%28u-6%29=%28-%28u-6%29%286%2Bu%29%29%2F%28u-6%29
%2836-u%5E2%29%2F%28u-6%29=-%286%2Bu%29
%2836-2x%29%2F%28sqrt%282x%29-6%29=-%286%2Bsqrt%282x%29%29
So now in the limit as x approaches 18.
lim%28x-%3E18%2C%2836-2x%29%2F%28sqrt%282x%29-6%29%29=-%286%2Bsqrt%282%2818%29%29%29
lim%28x-%3E18%2C%2836-2x%29%2F%28sqrt%282x%29-6%29%29=-%286%2Bsqrt%2836%29%29
lim%28x-%3E18%2C%2836-2x%29%2F%28sqrt%282x%29-6%29%29=-%286%2B6%29
lim%28x-%3E18%2C%2836-2x%29%2F%28sqrt%282x%29-6%29%29=-12