Question

lim (x tends to pi/2) sec7x/sec5x​

Answer 1

SOLUTION

TO EVALUATE

$\displaystyle \lim_{x \to \frac{\pi}{2} } \: \frac{ \sec 7x}{ \sec 5x}$

EVALUATION

SOLVE USING L' HOSPITAL RULE

$\displaystyle \lim_{x \to \frac{\pi}{2} } \: \frac{ \sec 7x}{ \sec 5x} \: \: \bigg( \sf{ \frac{ \infty }{ \infty } \: form \bigg)}$

$= \: \: \displaystyle \lim_{x \to \frac{\pi}{2} } \: \frac{ \frac{1}{ \cos 7x} }{ \frac{1}{ \cos 5x} } \:$

$= \: \displaystyle \lim_{x \to \frac{\pi}{2} } \: \frac{ \cos 5x}{ \cos 7x} \: \sf{ \: \bigg( \: \frac{0}{0} \: form \bigg)}$

$= \: \displaystyle \lim_{x \to \frac{\pi}{2} } \: \frac{ - 5 \sin 5x}{ - 7 \sin 7x} \:$

$= \: \displaystyle \: \frac{5}{7} \: \lim_{x \to \frac{\pi}{2} } \: \frac{ \sin 5x}{ \sin 7x} \:$

$= \: \displaystyle \: \frac{5}{7} \: \times \: \frac{ \sin \frac{5\pi}{2} }{ \sin \frac{7\pi}{2} } \:$

$= \: \displaystyle \: \frac{5}{7} \: \times \: \frac{ 1 }{ - 1 } \:$

$= \: \displaystyle \: - \frac{5}{7}$

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Answer 2

$\rm{\displaystyle\lim _{x\to \frac{\pi }{2}}\left(\frac{\sec \left(7x\right)}{\sec \left(5x\right)}\right)}$

• We are All Aware of Basic Trigonometry Identity  $\rm{sec(x)=\dfrac{1}{cos(x)}}$

$\rm{\displaystyle=\lim _{x\to \frac{\pi }{2}}\frac{\frac{1}{\cos \left(7x\right)}}{\sec \left(5x\right)}}$

$\rm{\displaystyle=\lim _{x\to \frac{\pi }{2}}\frac{\frac{1}{\cos \left(7x\right)}}{\frac{1}{\cos \left(5x\right)}}}$

$\rm{\displaystyle=\lim _{x\to \frac{\pi }{2}}\left(\frac{\cos \left(5x\right)}{\cos \left(7x\right)}\right)}$

• Apply L'Hopital's Rule

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$\underline{\underline{\mathbf{L'Hopital\:Theorem:}}}$

$\displaystyle\mathrm{For}\:\lim _{x\to a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right),\:\mathrm{if}\:\lim _{x\to a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{0}{0}\quad \mathrm{or}\quad \lim _{x\to \:a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\pm \infty }{\pm \infty },\:\mathrm{then}}$

$\displaystyle\bf{\lim _{x\to a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\lim _{x\to a}\left(\frac{f\:^'\left(x\right)}{g\:^'\left(x\right)}\right)}$

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$\rm{\displaystyle=\lim _{x\to \frac{\pi }{2}}\left(\frac{\left(\cos \left(5x\right)\right)^{'\:}}{\left(\cos \left(7x\right)\right)^{'\:}}\right)}$

$\rm{\displaystyle=\lim _{x\to \frac{\pi }{2}}\left(\frac{-\sin \left(5x\right)\cdot \:5}{-\sin \left(7x\right)\cdot \:7}\right)}$

$\rm{\displaystyle=\lim _{x\to \frac{\pi }{2}}\left(\frac{5\sin \left(5x\right)}{7\sin \left(7x\right)}\right)}$

• Plug in the value x = π/2

$\rm{\displaystyle=\frac{5\sin \left(5\cdot \frac{\pi }{2}\right)}{7\sin \left(7\cdot \frac{\pi }{2}\right)}}$

$\rm{\displaystyle=\frac{5}{7\sin \left(7\cdot \frac{\pi }{2}\right)}}$

$\rm{\displaystyle=\frac{5}{-7}}$

$\rm{\displaystyle=-\frac{5}{7}}$