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You can try the graphing approach, it will show that it goes towards 0.
You can solve this problem using l'Hospital's rule. You have to apply the rule twice, because after the first application you still have zero in the numerator and denominator. The second application is a little tricky because the second derivative of\[(\tan x-1)^2\]doesn't exactly leap off the page, but when you work through it you'll get a usable result. Post again if this isn't enough help for you to find the answer (which is not zero).
There is another approach. First, you can calculate lim_(x→π/4)(x-π/4)/(tanx-1). In doing this, you can use a trigonometric equation that tan(a+b)=(tana+tanb)/(1-tana*tanb) and convert tanx-1 to (tan(x-π/4))*(1+tanx). Therefore the function becomes (x-π/4)*cos(x-π/4)/sin(x-π/4)(1+tanx). When x→π/4, (x-π/4)→0 so lim (x-π/4)/ sin(x-π/4)=1 .All you need then is to plug π/4 into cos(x-π/4)/(1+tanx) which gets 1/2. Square it and you can get the answer 1/4. Hope it helps!
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Creeksider must have been half asleep not to notice that we can avoid the complexity of applying l'Hospital's rule twice by finding the limit of the square root and then squaring the limit. With that insight, though, it seems easier to solve with a single application of l'Hospital's rule than with a trig identity. Working with (x-π/4)/(tanx-1), the derivative of the numerator is 1 and the derivative of the denominator is sec²x, so we easily get cos²x. Then the answer is (cos²(π/4))² which is, as WPAN found, 1/4.