Question

find lim x tends to zero sinx/ x(1+ cosx)​

Answer 1

Given : lim x tends to zero sinx/ x(1+ cosx)​

$\lim_{x \to 0} \dfrac{\sin x}{x(1+\cos x)}$

To Find :  Evaluate the limit

Solution:

$\lim_{x \to 0} \dfrac{\sin x}{x(1+\cos x)}$

sinx  = 2sin(x/2)cos(x/2)

1 + cosx  = 2cos²(x/2)

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

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

$=\lim_{x \to 0} \dfrac{ \tan \frac{x}{2} }{x}$

$=\lim_{x \to 0} \dfrac{ \tan \frac{x}{2} }{2\frac{x}{2} }$

$= \frac{1}{2} \lim_{x \to 0} \dfrac{ \tan \frac{x}{2} }{\frac{x}{2} }$

 $\lim_{x \to 0} \dfrac{ \tan \frac{x}{2} }{ \frac{x}{2} } = 1$

=  1/2

Learn More:

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Answer 2

Step-by-step explanation: lim x-> 0 sin x/x ( 1 + cos x )

                                         multiplying and dividing x in numerator.

                               =>          lim x-> 0 (sin x/x *x)/x ( 1+cos x )

                              =>           lim x-> 0 (sin x / x ) / 1 + cos x

                              =>            lim x-> 0  1 / 1+cox x

                              =>             1 / 1 + cos 0

                              =>             1 / 1 + 1

                              =>              1/2

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