Limits and Derivatives — Maths STD 11 Science — Question
CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits and Derivatives1 Mark
Question
$\lim\limits_{\pi \rightarrow 0}\Big(\sin\text{mx}\cot\frac{\text{x}}{\sqrt{3}}\Big)=2$ then ___________.
✓
Answer
$\lim\limits_{\pi \rightarrow 0}\Big(\sin\text{mx}\cot\frac{\text{x}}{\sqrt{3}}\Big)=2$ then $\text{m}=\frac{2\sqrt{3}}{3}.$ Solution: Given $\lim\limits_{\pi \rightarrow 0}\Big(\sin\text{mx}\cot\frac{\text{x}}{\sqrt{3}}\Big)=2$ $=\lim\limits_{\pi \rightarrow 0}\frac{\sin\text{mx}}{\text{mx}}\times\text{mx}\lim\limits_{\text{x} \rightarrow 0}\Big(\cot\frac{\text{x}}{\sqrt{3}}\Big)=2$ $=1\times\text{mx}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{\tan\frac{\text{x}}{\sqrt{3}}}=2$ $=1\times\text{mx}\times\frac{\frac{\text{x}}{\sqrt{3}}}{\frac{\text{x}}{\sqrt{3}.\tan\frac{\text{x}}{\sqrt{3}}}}=2$ $=\frac{\text{mx}}{\frac{\text{x}}{\sqrt{3}}}(1)=2$ $=\sqrt{2}\text{m}=2$ $\text{m}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$ Hence, the value of the filler is $\frac{2\sqrt{3}}{3}.$
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