Question

$\lim_{x \to \ \frac{ \pi }{4} } \frac{1-tanx}{1- \sqrt{2}sinx }$

Answer 1

$\lim\limits_{x\to\frac{\pi}{4}}\frac{1-tanx}{1-\sqrt2sinx}=\left[\frac{0}{0}\right]\\\\Use\ the\ l'Hopital's\ rule:\\\\f(x)=1-tanx\to f'(x)=-\frac{1}{cos^2x}\\\\g(x)=1-\sqrt2sinx\to g'(x)=-\sqrt2cosx\\\\\lim\limits_{x\to\frac{\pi}{4}}\frac{1-tanx}{1-\sqrt2sinx}=\lim\limits_{x\to\frac{\pi}{4}}\frac{-\frac{1}{cos^2x}}{-\sqrt2cosx}=\lim\limits_{x\to\frac{\pi}{4}}\frac{1}{\sqrt2cos^3x}=\frac{1}{\sqrt2\cdot\left(\frac{\sqrt2}{2}\right)^3}$

$=\frac{1}{\sqrt2\cdot\frac{2\sqrt2}{8}}=\frac{1}{\frac{4}{8}}=\frac{1}{\frac{1}{2}}=2$

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$The\ l'Hospital's\ rule:\\\\if\\\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}g(x)=0\ or\pm\infty\ and\ \lim\limits_{x\to a}\frac{f'(x)}{g'(x)}\ exist\ where\ g'(x)\neq0\\\\then\\\lim\limits_{x\to a}\frac{f(x)}{g(x)}=\lim\limits_{x\to a}\frac{f'(x)}{g'(x)}$

Answer 2

let x = π/4 + Δx    So our limit is as Δx -> 0
tan π/4 = 1  sin 45 = 1/√2
numerator =  1 - tan (π/4 + Δx )  =  1  - [ (tan 45 + tan Δx)/(1 - tan 45  tan Δx) ]
             = 1 -  [ (1 + tan Δx)/ (1-tan Δx) ]  =  - 2  tan Δx  / (1 - tan Δx)
Numerator / Δx = - 2 ( tan Δx  / Δx ) * 1/(1 - tan Δx)
Lim        Numerator / Δx  =  - 2       as Lim (tan Δx / Δx) = 1  and  Lim tan Δx = 0
Δx->0                                                Δx->0                              Δx->0
Denom = 1 - √2 sin (45+Δx) = 1 - Cos Δx - Sin Δx 
           = 2 sin² Δx/2 - 2 Sin Δx/2 Cos Δx/2  = 2 Sin Δx/2 [ sin Δx/2 - cos Δx/2]
Denom / Δx  = ( Sin Δx/2  / Δx/2)    [ sin Δx/2  - cos Δx/2 ]
Lim          Denom / Δx    =  -1        as Lim sin Δx/2  / Δx/2  = 1
Δx -> 0                                             Δx/2 -> 0
               Lim  Δx/2 ->0  of Sin Δx/2 =0        and for cos ,  it is 1.
So numerator / Denominator as Lim Δx->0  ,  -2 / -1  = 2