Question

limit x tends to π/4 [(√2-cos x-sin x)/(4x-π)^2]
evaluate the above limit​

Answer 1

Answer

Using L'Hopital's rule

L'HOPITAL'S rule

Since evaluating the limits of numerator and denominator would result in an intermediate form.

$\lim_{x→ \frac{\pi}{4} }( \frac{ \frac{dx}{dy}( \sqrt{2} - \cos(x) - \sin(x) )}{ \frac{dx}{dy} {(4x - \pi)}^{2} } )$

Calculate the derivatives

$\lim_{x→ \frac{\pi}{4} }( \frac{ \sin(x) - \cos(x) }{32x - 8\pi} )$

Use L'Hopital's rule

$\lim_{x→ \frac{\pi}{4} }( \frac{ \frac{ dx}{ dy}( \sin(x) - \cos(x) ) }{ \frac{dx}{dy} (32x - 8\pi)} )$

Calculate the derivatives

$\lim_{x→ \frac{\pi}{4} }( \frac{ \cos(x) + \sin(x) }{32} )$

Evaluate the limit

$= \frac{ \cos( \frac{ {\pi}^{c} }{4} ) + \sin( \frac{ {\pi}^{c} }{4} ) }{32}$

We know that

$\frac{ {\pi}^{c} }{4} = 45°$

$= \frac{ \cos(45°) + \sin(45°) }{32}$

Using

cos45° = 1/√2

sin45° = 1/√2

$= \frac{ \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } }{32}$

$= \frac{ \frac{ \sqrt{2} + \sqrt{2} }{ \sqrt{4} } }{32}$

$= \frac{ \frac{2 \sqrt{2} }{2} }{32}$

$= \dfrac{ \sqrt{2} }{32}$