Hi

Please helpe me to demonstrate that lim x->0 [ln(1+x)+ln(1-x)]/[cos(x)-1]= 2

Use L'hopital's Rule.

i don't know how ...please helpe me .

To demonstrate the limit, we can use L'Hôpital's rule which states that if we have an indeterminate form, such as 0/0 or ∞/∞, we can differentiate the numerator and denominator repeatedly until we get a form that is no longer indeterminate.

Let's begin by differentiating the numerator and denominator with respect to x:

Numerator:
d/dx [ln(1+x) + ln(1-x)]
= 1/(1+x) - 1/(1-x)
= (1-x - 1-x) / (1+x)(1-x)
= -2x / (1+x)(1-x)

Denominator:
d/dx [cos(x) - 1]
= -sin(x)

Now, we rewrite the original limit expression using the derivatives:

lim x->0 [ln(1+x) + ln(1-x)] / [cos(x) - 1]
= lim x->0 [-2x / (1+x)(1-x)] / -sin(x)
= lim x->0 2x / [(1+x)(1-x)] * (1 / sin(x))

Next, we can factor out the common terms:

= 2 / [(1+x)(1-x)] * lim x->0 x / sin(x)

Now, we can evaluate the limit as x approaches 0. The limit of x/sin(x) as x approaches 0 is equal to 1. Therefore, we have:

= 2 / [(1+0)(1-0)] * 1
= 2 / (1 * 1)
= 2

Thus, we have proven that lim x->0 [ln(1+x)+ln(1-x)]/[cos(x)-1] = 2.