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Solving: lim⁡x→π2(sin⁡(x)sin⁡(x)2+9cos⁡(x)22cos⁡(x))\lim_{x\to{\frac{\pi }{2}}}\left(\frac{\sin\left(x\right)\sqrt{\sin\left(x\right)^2+9\cos\left(x\right)^2}}{2\cos\left(x\right)}\right)
Solve instead: lim⁡x→π2(sin(x)sin2x+9cos2t2cosx)\lim_{x\to\frac{\pi}{2}}\left(\frac{sin\left(x\right)\sqrt{sin^2x+9cos^2t}}{2cosx}\right)
f(x)=(sin(x)(sin(x)^2+9cos(x)^2)^(1/2))/(2cos(x))−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

lim⁡x→π2(sin(x)sin2x+9cos2t2cosx)\lim_{x\to\frac{\pi}{2}}\left(\frac{sin\left(x\right)\sqrt{sin^2x+9cos^2t}}{2cosx}\right)

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Final answer to the exercise

The limit does not exist

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limx→π2 (sin(x)√sin2x+9cos2t2cosx )
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