I need help calculating this limit:limx→πcos⁡2x+2(π−x)sin⁡x−1(cos⁡2x+cos⁡x)2I tried to use L'Hospital's rule, but didn't get any result. Can you help me please?

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I need help calculating this limit:limx→πcos⁡2x+2(π−x)sin⁡x−1(cos⁡2x+cos⁡x)2I tried to use L&#039;Hospital&#039;s rule, but didn&#039;t get any result. Can you help me please?

Jazlyn Mitchell · Accepted Answer

Let x=t+πThus,limx→πcos⁡2x+2(π−x)sin⁡x−1(cos⁡2x+cos⁡x)2=limt→0cos⁡2t+2tsin⁡t−1(cos⁡2t−cos⁡t)2==limt→02tsin⁡t−2sin2t(1−cos⁡t)2(2cos⁡t+1)2=29limt→0tsin⁡t−sin2t4sin4t2==89limt→0tsin⁡t−sin2tt4=89limt→0t−sin⁡tt3==89limt→01−cos⁡{t}3t2=89limt→02sin2t23t2=89limt→02⋅t243t2=427

Charlie Haley · Accepted Answer

Hint: breaks it like thislimx→πcos⁡2x+2(π−x)sin⁡x−1(cos⁡2x+cos⁡x)2=limx→π(cos⁡2x−1(π−x)2+2sin⁡x(π−x))⋅limx→π(π−xcos⁡2x+cos⁡x)2Then use the definition of derivative of a function at x=π

I need help calculating this limit: \(\displaystyle\lim_{{{x}\to\pi}}{\frac{{{\cos{{2}}}{x}+{2}{\left(\pi-{x}\right)}{\sin{{x}}}-{1}}}{{{\left({\cos{{2}}}{x}+{\cos{{x}}}\right)}^{{2}}}}}\) I tried

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