Question 273224
{{{lim(x->1, (x-1)/(sqrt(x)-1))}}}
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(x)-1)}}} by its conjugate, {{{sqrt(x)+1)}}}. From the pattern {{{(a+b)(a-b) = a^2 -b^2}}} we know that when you multiply conjugates you get the difference of the squares of the two terms. Let's see how this helps:
{{{lim(x->1, ((x-1)/(sqrt(x)-1))((sqrt(x)+1)/(sqrt(x)+1)))}}}
which simplifies to:
{{{lim(x->1, ((x-1)(sqrt(x)+1))/(x - 1))}}}
A fine but important point is that in this limit x approaches 1 <i>but is never actually equal to 1!</i> This is important because if x is 1, x-1 is zero and we <b>cannot</b> cancel 0/0. But since x is never 1, x-1 is never 0 and we can cancel the (x-1)'s:
{{{lim(x->1, sqrt(x)+1)}}}
This limit is simple to find. I'll leave it up to you to finish.<br>
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(x->1, (sqrt(x)-1)/(x-1))}}}
then your answer will be "upside down", too (which matches the answer key's answer of 1/2).