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  • Find <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{(1+3 x)^{1 / 3}-\sin (x)-1}{1-\cos (x)}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%200%7D

Find \mathrm{\lim _{x \rightarrow 0} \frac{(1+3 x)^{1 / 3}-\sin (x)-1}{1-\cos (x)}}
 

Option: 1

-2


Option: 2

0


Option: 3

1


Option: 4

5


Answers (1)

best_answer

\mathrm{\frac{(1+3 x)^{\frac{1}{3}}-1-(\sin x)}{1-(\cos x)} =\frac{\left((1+3 x)^{\frac{1}{3}}-1-x\right)-(\sin x-x)}{1-(\cos x)} }

                                             \mathrm{ =\frac{9 \times\left(\frac{(1+3 x)^{\frac{1}{3}}-1-\frac{1}{3}(3 x)}{(3 x)^2}\right)-\left(\frac{\sin x-x}{x^2}\right)}{\left(\frac{1-\cos x}{x^2}\right)} }

Using standard limits
\mathrm{ \lim _{u \rightarrow 0} \frac{(1+u)^{\frac{1}{3}}-1-\frac{1}{3} u}{u^2}=-\frac{1}{9}, \quad \text { and } \quad \lim _{x \rightarrow 0} \frac{\sin x-x}{x^2}=0, \quad \text { and } \quad \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2} }
the required limit follows

\mathrm{ \lim _{x \rightarrow 0} \frac{(1+3 x)^{\frac{1}{3}}-1-(\sin x)}{1-(\cos x)} =\frac{9 \times \lim _{x \rightarrow 0}\left(\frac{(1+3 x)^{\frac{1}{3}}-1-\frac{1}{3}(3 x)}{(3 x)^2}\right)-\lim _{x \rightarrow 0}\left(\frac{\sin x-x}{x^2}\right)}{\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^2}\right)} }

                                                      \mathrm{ =\frac{9 \times\left(-\frac{1}{9}\right)-(0)}{\left(\frac{1}{2}\right)}=-2 . }

Hence option 1 is correct.
 

Posted by

shivangi.bhatnagar

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