Putting z=x-pi/4 show that limit of (1-tanx)/(1-sqrt2 sinx)=2 as x approaches to pi/4?

1 Answer
Dec 12, 2017

x = z+pi/4x=z+π4

tanx = tan(z+pi/4)tanx=tan(z+π4)

= frac{tanz+tan frac{pi}{4}}{1-tanz tan frac{pi}{4}}=tanz+tanπ41tanztanπ4

= frac{tanz + 1}{1 - tanz (1)}=tanz+11tanz(1)

= frac{1+ tanz}{1 - tanz}=1+tanz1tanz

sinx = sin(z+pi/4)sinx=sin(z+π4)

= sinz cos frac{pi}{4} + cosz sin frac{pi}{4}=sinzcosπ4+coszsinπ4

= 1/sqrt2 (sinz + cosz)=12(sinz+cosz)

lim_{x->pi/4} (1 - tanx)/(1-sqrt2 sinx)

= lim_{z->0} (1 - (1+ tanz)/(1 - tanz))/(1-sqrt2 1/sqrt2 (sinz + cosz))

= lim_{z->0} ((1 -tan z - (1+ tanz))/(1 - tanz))/(1- sinz - cosz)

= lim_{z->0} (-2tanz)/((1- sinz - cosz)(1 - tanz))

(Assuming limit exists)

= lim_{z->0} (-2tanz)/(1- sinz - cosz)

= 2 lim_{z->0} (tanz)/(sinz + cosz - 1)

= 2 lim_{z->0} ((sinz + cosz - 1)/tanz)^(-1)

= 2 lim_{z->0} (sinz/tanz - (1-cosz)/tanz)^(-1)

= 2 lim_{z->0} (cosz - (1-cos^2z)/(tanz(1+cosz)))^(-1)

= 2 lim_{z->0} (1 - (sin^2z)/(tanz(1+cosz)))^(-1)

= 2 lim_{z->0} (1 - (sinzcosz)/(1+cosz))^(-1)

= 2 lim_{z->0} (1 - ((0)(1))/(1+1))^(-1)

= 2