29@Pranav prince....looks like u need to apply L'Hospital rule twice in this case...btw solving the problem...will post the solution in a while'
11 Answers
1didnt get it right in latex! http://targetiit.com/latex/editor.php?target=message&html
291i did that twice but didnt get the answer given in the back of the book!
29\lim_{x\pi /4}\frac{(sinx + cosx)^{3} - 2\sqrt{2}}{1-sin2x} = \lim_{x\rightarrow \pi /4}\frac{3(sinx + cosx)^{2}(cosx-sinx)}{-2cos2x} =
= \lim_{x\rightarrow \pi /4} [\frac{6(sinx+cosx)(cosx- sinx)^{2} + 3 (sinx+cosx)^{2}(-sinx-cosx)}{4sin2x} = \frac{-3\sqrt{2}}{2}
