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

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

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Answer 1 · 2017-12-12T13:46:15

$x = z + \frac{π}{4}$ $x = z+pi/4$

$tan x = tan (z + \frac{π}{4})$ $tanx = tan(z+pi/4)$

$= \frac{tan z + tan \frac{π}{4}}{1 - tan z tan \frac{π}{4}}$ $= frac{tanz+tan frac{pi}{4}}{1-tanz tan frac{pi}{4}}$

$= \frac{tan z + 1}{1 - tan z (1)}$ $= frac{tanz + 1}{1 - tanz (1)}$

$= \frac{1 + tan z}{1 - tan z}$ $= frac{1+ tanz}{1 - tanz}$

$sin x = sin (z + \frac{π}{4})$ $sinx = sin(z+pi/4)$

$= sin z cos \frac{π}{4} + cos z sin \frac{π}{4}$ $= sinz cos frac{pi}{4} + cosz sin frac{pi}{4}$

$= \frac{1}{\sqrt{2}} (sin z + cos z)$ $= 1/sqrt2 (sinz + cosz)$

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

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

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

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

(Assuming limit exists)

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

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

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

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

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

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

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

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

$= 2$

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

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