Question

lim x tends to 0 log sin 2x/log sinx​

Answer 1

Given:

$\lim_{x \to 0} \frac{log(sin(2x))}{log(sin(x))}$

To find:

Limit of given expression.

Solution:

Left hand limit:

Since we have an indeterminate form of type ∞/∞, we can apply the L'Hopital's rule.

$\lim_{x \to 0^{-} } \frac{log(sin(2x))}{log(sin(x))} = \lim_{x \to 0^{-}} \frac{\frac{d}{dx} (log(sin(2x)))}{\frac{d}{dx} (log(sin(x)))}$

$\lim_{x \to 0^{-}} \frac{\frac{d}{dx} (log(sin(2x)))}{\frac{d}{dx} (log(sin(x)))} = \lim_{x \to 0^{-}}\frac{2sin(x)cos(2x)}{sin(2x)cos(x)}$

Simplify,

$\lim_{x \to 0^{-}}\frac{2sin(x)cos(2x)}{sin(2x)cos(x)} = \lim_{x \to 0^{-}}(2 - \frac{1}{cos^{2}(x)} )$

Substitute the variable with the value

$\lim_{x \to 0^{-}}(2 - \frac{1}{cos^{2}(x)} ) = 1$

Therefore,

$\lim_{x \to 0^{-}} \frac{log(sin(2x))}{log(sin(x))} = 1$

Right hand limit:

Since we have an indeterminate form of type ∞/∞, we can apply the L'Hopital's rule.

$\lim_{x \to 0^{+} } \frac{log(sin(2x))}{log(sin(x))} = \lim_{x \to 0^{+}} \frac{\frac{d}{dx} (log(sin(2x)))}{\frac{d}{dx} (log(sin(x)))}$

$\lim_{x \to 0^{+}} \frac{\frac{d}{dx} (log(sin(2x)))}{\frac{d}{dx} (log(sin(x)))} = \lim_{x \to 0^{+}}\frac{2sin(x)cos(2x)}{sin(2x)cos(x)}$

Simplify,

$\lim_{x \to 0^{+}}\frac{2sin(x)cos(2x)}{sin(2x)cos(x)} = \lim_{x \to 0^{+}}(2 - \frac{1}{cos^{2}(x)} )$

Substitute the variable with the value

$\lim_{x \to 0^{+}}(2 - \frac{1}{cos^{2}(x)} ) = 1$

Therefore,

$\lim_{x \to 0^{+}} \frac{log(sin(2x))}{log(sin(x))} = 1$

Since, LHS = RHS = 1.

Therefore, answer is $\lim_{x \to 0} \frac{log(sin(2x))}{log(sin(x))} = 1$

Answer 2

Step-by-step explanation:

lim

x→0

+

sin(2x)cos(x)

2sin(x)cos(2x)

=lim

x→0

+

(2−

cos

2

(x)

1

)