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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

-\sqrt{3} / 16


Option: 2

\sqrt{5} / 16


Option: 3

0


Option: 4

None


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 \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 simplifies 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}.

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