Question

lim x tends to (pi/4)
$1 - \tan(x) \div 1 - \sqrt{2} \sin(x)$


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Answer 1

$\huge\underline\blue{AnSweR}$

Setting π4−x=2h

tanx=tan(π4−2h)=1−tan2h1+tan2h ⟹1−tanx=2tan2h1+tan2h

and sinx=sin(π4−2h)=cos2h−sin2h2–√

limx→π41−tanx1−2–√sinx=limh→02tan2h(1+tan2h)(1−cos2h+sin2h)

=limh→0sin2h1−cos2h+sin2h⋅limh→02cos2h(1+tan2h)

Now the second limit is easy,

For the first using Double-Angle Formulas, sin2h1−cos2h+sin2h=2sinhcosh1−(1−2sin2h)+2sinhcosh =coshsinh+cosh if sinh≠0 which is true as h→0....

Hope it helps you!!