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  • Evaluate <img alt="\mathrm{\lim _{x \rightarrow 4}\left(\frac{\frac{\pi}{6}-\arcsin \left(\frac{\sqrt{x}}{4}\right)}{\sqrt[3]{2 x-7}-1}\right)}" src="https://entrancecorner.oncodecogs.com/gif.

Evaluate \mathrm{\lim _{x \rightarrow 4}\left(\frac{\frac{\pi}{6}-\arcsin \left(\frac{\sqrt{x}}{4}\right)}{\sqrt[3]{2 x-7}-1}\right)}

Option: 1

\mathrm{-\frac{\sqrt{3}}{16}}


Option: 2

\mathrm{\frac{\sqrt{3}}{16}}


Option: 3

\mathrm{\frac{\sqrt{5}}{16}}


Option: 4

\mathrm{0}


Answers (1)

best_answer

Well, let's put \mathrm{u=\arcsin (\sqrt{x} / 4)} so that as \mathrm{x\rightarrow 4} we have \mathrm{u\rightarrow \frac{\pi}{6}} and \mathrm{x=16\: sin^{2}u}. The expression under limit can be written as

\mathrm{\frac{1}{16} \cdot \frac{\pi / 6-u}{1 / 2-\sin u} \cdot \frac{1}{1 / 2+\sin u} \cdot \frac{4-x}{\sqrt[3]{2 x-7}-1}}

The desired limit is then equal to 

\mathrm{\frac{1}{16} \cdot \frac{1}{\cos (\pi / 6)} \cdot 1 \cdot \lim _{x \rightarrow 4} \frac{4-x}{\sqrt[3]{2 x-7}-1}}

The above simplifis to

\mathrm{\frac{1}{8 \sqrt{3}} \lim _{t \rightarrow 1} \frac{1-t^3}{2(t-1)}}

using substitution \mathrm{t=\sqrt[3]{2 x-7}},Thus the desired limit is \mathrm{-\sqrt{3} / 16}

Posted by

Irshad Anwar

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