SOLUTION: Evaluate each limit. Rationalize the numerator by multiplying both numerator and denominator √x+1. lim x→1 x-1/√x-1

Algebra ->  Radicals -> SOLUTION: Evaluate each limit. Rationalize the numerator by multiplying both numerator and denominator √x+1. lim x→1 x-1/√x-1      Log On


   

Question 273224: Evaluate each limit. Rationalize the numerator by multiplying both numerator and denominator √x+1.
lim
x→1
x-1/√x-1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
lim%28x-%3E1%2C+%28x-1%29%2F%28sqrt%28x%29-1%29%29
The "trick" you will see here is used a lot to find limits where the denominator appears to approach zero. What we will do is multiply sqrt%28x%29-1%29 by its conjugate, sqrt%28x%29%2B1%29. From the pattern %28a%2Bb%29%28a-b%29+=+a%5E2+-b%5E2 we know that when you multiply conjugates you get the difference of the squares of the two terms. Let's see how this helps:

which simplifies to:
lim%28x-%3E1%2C+%28%28x-1%29%28sqrt%28x%29%2B1%29%29%2F%28x+-+1%29%29
A fine but important point is that in this limit x approaches 1 but is never actually equal to 1! This is important because if x is 1, x-1 is zero and we cannot cancel 0/0. But since x is never 1, x-1 is never 0 and we can cancel the (x-1)'s:
lim%28x-%3E1%2C+sqrt%28x%29%2B1%29
This limit is simple to find. I'll leave it up to you to finish.

Responding to your message: This is all correct as long as the problem you posted is correct. Of course if the problem is actually:
lim%28x-%3E1%2C+%28sqrt%28x%29-1%29%2F%28x-1%29%29
then your answer will be "upside down", too (which matches the answer key's answer of 1/2).