Question

lim x approaches 1 of ((square root of x)+1)/(x-1)

Answer 1

Youd have to rationalize the numerator by multiplying the top and bottom by the conjugate.

Answer 2

\[\frac{ \sqrt{x}+1 }{ x-1 }*\frac{ \sqrt{x}-1 }{ \sqrt{x} -1} \]

Answer 3

if I plug in 1 in the new equation, wouldn't I get another undefined answer?

Answer 4

Well, let's multiply it out and see ^_^

Answer 5

\[\frac{ x-1 }{ \sqrt{x ^{3}}-x-\sqrt{x}+1 }\]
\[\frac{ 1-1 }{ \sqrt{1^{3}}-1-\sqrt{1}+1 } =\frac{ 0 }{ 0 } \]
So yes, still undefined. Now there would be one last way to check with a calc 2 method, but in the end it does appear your answer is does not exist :P

Answer 6

I got this wrong on as quiz by doing the method above of rationalizing the numerator and getting undefined in the denominator, i think the teacher wants the calc 2 method, How did you do it?

Answer 7

Well even with the calc 2 method you still get no limit. The calc 2 method requires us to have an indeterminant form, mainly 0/0 or infinity/infinity. Now we did not originally have this, but rationalizing the numerator gave us this option. So since we have 0/0, we can try to use l'hopitals rule, which states that the limit of 2 functions divided by each other is the same as the derivativesof two functions divided by each other. So basically, we take the derivative of the numerator, SEPARATELY, and divided it by the derivative of the denominator SEPARATELY.

Answer 8

The reason I stress separately is because a lot of students try to take the derivative of the whole fraction, which is incorrect.

Answer 9

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Answer 10

Thank you Psymon!

Answer 11

Yep, np :3