Question

the value of lim x tends to 0 sin[log (1+x)]/log(1+sinx)

Answer 1

multiply divide with log(1+x)
sinΦ/Φ=1
log(1+x)/log(1+sinΦ)
by L. h. rule
1/1+x/cosΦ/1+sinΦ
1+sinx/(1+x)(cosx )
limt x to 0
1+0/(1+0)(1)
1 ans
hope it helped
tell me if it is correct

Answer 2

Answer:

The value of given limit is

$lim _{x - > 0} \frac{sin(log(1 + x))}{log(1 + sinx)}$

Step-by-step explanation:

The given limit is

$lim _{x - > 0} \frac{sin(log(1 + x))}{log(1 + sinx)}$

Applying the value of given limit we get the value as 0/0 which is an indeterminate form.Hence we apply L-hospital rule by differentiating numerator and denominator we get

$lim _{x - > 0} \frac{cos(log(1 + x)). \frac{1}{1 + x} }{ \frac{1}{1 + sinx} .cosx}$

Now substituting the value of x=0,we get the limit as

$lim _{x - > 0} \frac{cos(log1).1}{ \frac{1}{1 + 0}.cos0 } \\ = 1$

Therefore,the value of given limit is

$lim _{x - > 0} \frac{sin(log(1 + x))}{log(1 + sinx)} = 1$

#SPJ3