Question

Lt log sec2x/logsecx
X->0
Plus Help. It's urgent!!!

Answer 1

Answer:

$\large\boxed{\sf{2}}$

Step-by-step explanation:

To evaluate the given limit,

$\displaystyle\lim_{x\to0} \dfrac{ log( \sec(2x) ) }{ log( \sec(x) ) }$

But, we know that,

• $\sec( \alpha ) = \dfrac{1}{ \cos( \alpha ) }$

Therefore, we will get,

$= \displaystyle\lim_{x\to0} \dfrac{ log( \frac{1}{ \cos(2x) } ) }{ log( \frac{1}{ \cos (x) } ) }$

But, we know that,

• $log( \dfrac{x}{y} ) = log( x ) - log(y)$

Therefore, we will get,

$= \displaystyle\lim_{x\to0} \dfrac{ log(1) - log( \cos(2x) ) }{ log(1) - log( \cos(x) ) }$

But, we know that,

• $log(1) = 0$

Therefore, we will get,

$= \displaystyle\lim_{x\to0} \dfrac{ - log( \cos(2x) ) }{ - log( \cos(x) ) } \\ \\ = \displaystyle\lim_{x\to0} \dfrac{ log( \cos(2x) ) }{ log( \cos(x) ) }$

Now, applying L Hospitals rule, i.e., differentiatiating both numerator and denominator separately, we will get,

$= \displaystyle\lim_{x\to0} \dfrac{ \dfrac{1}{ \cos(2x) } \times 2 \cos(x) \cancel{( - \sin(x)) } }{ \dfrac{1}{ \cos(x) } \times \cancel{( - \sin(x))} } \\ \\ = \displaystyle\lim_{x\to0} \dfrac{2 { \cos }^{2}x }{ \cos2x}$

Now, substituting the limits, we get,

$= \dfrac{2 { \cos }^{2} 0 \degree}{ \cos0 \degree } \\ \\ = \dfrac{2 \times 1}{1} \\ \\ = 2$

Hence, the required value is 2