Question

limit x tends to zero log cosx /sin^2x

Answer 1

HEY THERE,

lim    [log(cos x)/sinx]
x→0

Applying L Hospital's rule:--

=lim   [-sin x/sin x*2cos²x]
  x→0

=lim   [-1/2cos²x]
  x→0 

=-1/2

HOPE IT HELPS
  

Answer 2

$\lim_{x \to 0} \dfrac{\log \cos x}{\sin^2 x}=\dfrac{-1}{2}$

Step-by-step explanation:

We have,

$\lim_{x \to 0} \dfrac{\log \cos x}{\sin^2 x}$

To find, $\lim_{x \to 0} \dfrac{\log \cos x}{\sin^2 x}=?$

∴$\lim_{x \to 0} \dfrac{\log \cos x}{\sin^2 x}$ ($\dfrac{0}{0}$ form)

Applying L'Hopital's rule,

$\lim_{x \to 0} \dfrac{\dfrac{-\sin x}{\cos x} }{2\sin x\cos x}$

$=\lim_{x \to 0} \dfrac{-1}{2\cos^2 x}$

Put x = 0, we get

$=\dfrac{-1}{2\cos^2 0}$

$=\dfrac{-1}{2\times 1}$

[ ∵ cos 0° = 1]

$=\dfrac{-1}{2}$

Hence, $\lim_{x \to 0} \dfrac{\log \cos x}{\sin^2 x}=\dfrac{-1}{2}$