Finding the limx→01−cos⁡xsin⁡xln⁡(1+x) using Taylors

Question

Samantha Brown · Accepted Answer

What you did is absolutely correct and i hope that i can start from there

lim_{x \to 0} \frac{1 - \cos x}{\sin x \ln (1 + x)} = lim_{x \to 0} \frac{\frac{x^{2}}{2} - \frac{x^{4}}{24}}{(x - \frac{x^{3}}{6}) (x - \frac{x^{2}}{2} + \frac{x^{3}}{3})}

now take

x^{2}

common in both numerator and denominator(one x from each braces in the denominator) and then check the coefficient of constant in both numerator and denominator which gives you the value of limit (since all other powers of x tend to zero as x tends to zero) and in this case the value of limit is

lim_{x \to 0} \frac{1 - \cos x}{\sin x \ln (1 + x)} = \frac{12}{}

puhnut1m · Accepted Answer

Do some asymptotic analysis, it greatly simplifies things: 1−cos⁡x∼0−x22 sin⁡x∼0x,  ln⁡(1+x∼0x.) Therefore 1−cos⁡xsin⁡xln⁡(1+x)∼0−x22x⋅x=−12

Finding the \lim_{x \to 0} \frac{1-\cos x}{\sin x \ln(1+x)} using

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