document.write( "Question 273224: Evaluate each limit. Rationalize the numerator by multiplying both numerator and denominator √x+1.\r
\n" ); document.write( "\n" ); document.write( "lim
\n" ); document.write( "x→1\r
\n" ); document.write( "\n" ); document.write( "x-1/√x-1 \n" ); document.write( "
Algebra.Com's Answer #200200 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"lim%28x-%3E1%2C+%28x-1%29%2F%28sqrt%28x%29-1%29%29\"
\n" ); document.write( "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:
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\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"lim%28x-%3E1%2C+%28%28x-1%29%28sqrt%28x%29%2B1%29%29%2F%28x+-+1%29%29\"
\n" ); document.write( "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:
\n" ); document.write( "\"lim%28x-%3E1%2C+sqrt%28x%29%2B1%29\"
\n" ); document.write( "This limit is simple to find. I'll leave it up to you to finish.

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