{"id":20201,"date":"2024-04-15T05:49:44","date_gmt":"2024-04-15T05:49:44","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20201"},"modified":"2024-04-15T05:49:44","modified_gmt":"2024-04-15T05:49:44","slug":"the-mathop-lim-limits_textx-to-0-fracsin-left-frac23textx-righttextx-is-a-frac23-b-1-c-frac32-d-infty","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-mathop-lim-limits_textx-to-0-fracsin-left-frac23textx-righttextx-is-a-frac23-b-1-c-frac32-d-infty\/","title":{"rendered":"The \\[\\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{\\sin \\left[ {\\frac{2}{3}{\\text{x}}} \\right]}}{{\\text{x}}}\\] is A. \\[\\frac{2}{3}\\] B. 1 C. \\[\\frac{3}{2}\\] D. \\[\\infty \\]"},"content":{"rendered":"

[amp_mcq option1=”\\[\\frac{2}{3}\\]” option2=”1″ option3=”\\[\\frac{3}{2}\\]” option4=”\\[\\infty \\]” correct=”option3″]<\/p>\n

The correct answer is $\\boxed{\\frac{2}{3}}$.<\/p>\n

The limit $\\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{\\sin \\left[ {\\frac{2}{3}{\\text{x}}} \\right]}}{{\\text{x}}}$ can be evaluated using L’H\u00c3\u00b4pital’s rule. L’H\u00c3\u00b4pital’s rule states that if $\\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{f\\left( {\\text{x}} \\right)}}{{g\\left( {\\text{x}} \\right)}}$ exists and $\\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{f’\\left( {\\text{x}} \\right)}}{{g’\\left( {\\text{x}} \\right)}}$ also exists, then $\\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{f\\left( {\\text{x}} \\right)}}{{g\\left( {\\text{x}} \\right)}} = \\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{f’\\left( {\\text{x}} \\right)}}{{g’\\left( {\\text{x}} \\right)}}$.<\/p>\n

In this case, we have $f\\left( {\\text{x}} \\right) = \\sin \\left[ {\\frac{2}{3}{\\text{x}}} \\right]$ and $g\\left( {\\text{x}} \\right) = {\\text{x}}$. Therefore, we have<\/p>\n

$$\\begin{align}
\n\\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{\\sin \\left[ {\\frac{2}{3}{\\text{x}}} \\right]}}{{\\text{x}}} &= \\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{f’\\left( {\\text{x}} \\right)}}{{g’\\left( {\\text{x}} \\right)}} \\\\
\n&= \\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{2\\cos \\left[ {\\frac{2}{3}{\\text{x}}} \\right] \\cdot \\frac{2}{3}}}{{1}} \\\\
\n&= \\frac{2}{3}.
\n\\end{align<\/em>}$$<\/p>\n

Therefore, the limit $\\mathop {\\lim }\\limits_{{\\text{x}} \\to 0} \\frac{{\\sin \\left[ {\\frac{2}{3}{\\text{x}}} \\right]}}{{\\text{x}}}$ is $\\boxed{\\frac{2}{3}}$.<\/p>\n

The other options are incorrect because they do not represent the correct value of the limit.<\/p>\n","protected":false},"excerpt":{"rendered":"

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