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Finding the \mathrm{ \lim _{x \rightarrow 0} \frac{1-\cos (x)}{\sin (x) \ln (1+x)}}

Option: 1

\frac{1}{2}


Option: 2

0


Option: 3

1


Option: 4

2


Answers (1)

best_answer

\mathrm{ \lim _{x \rightarrow o} \frac{1-\cos x}{\sin (x) \ln (1+x)}=\lim _{x \rightarrow 0} \frac{\frac{x^2}{2}-\frac{x^4}{24}}{\left(x-\frac{x^3}{6}\right)\left(x-\frac{x^2}{2}+\frac{x^3}{3}\right)} }

now take \mathrm{ x^2 } common in both numerator and denominator(one \mathrm{ 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 \mathrm{ x } tend to zero as \mathrm{ x } tends to zero) and in this case the value of limit is

\mathrm{ \lim _{x \rightarrow o} \frac{1-\cos x}{\sin (x) \ln (1+x)}=\frac{1}{2} }
 

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manish

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