[amp_mcq option1=”1″ option2=”0″ option3=”\\[\\frac{1}{2}\\]” option4=”2″ correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\frac{1}{2}}$.<\/p>\n
We can use the squeeze theorem to solve this problem. The squeeze theorem states that if $f(x) \\leq g(x) \\leq h(x)$ for all $x$ in a given interval, and $\\lim_{x \\to a} f(x) = \\lim_{x \\to a} h(x)$, then $\\lim_{x \\to a} g(x)$ also exists and is equal to the same value.<\/p>\n
In this case, we have $2 – \\frac{{{{\\text{x}}^2}}}{3} < \\frac{{{\\text{x}}\\sin {\\text{x}}}}{{1 – \\cos {\\text{x}}}} < 2$ for all $x$ close to 0. We also know that $\\lim_{x \\to 0} 2 – \\frac{{{{\\text{x}}^2}}}{3} = 1$ and $\\lim_{x \\to 0} 2 = 2$. Therefore, by the squeeze theorem, we have $\\lim_{x \\to 0} \\frac{{{\\text{x}}\\sin {\\text{x}}}}{{1 – \\cos {\\text{x}}}} = \\frac{1}{2}$.<\/p>\n
We can also solve this problem by using L’H\u00c3\u00b4pital’s rule. L’H\u00c3\u00b4pital’s rule states that if $\\lim_{x \\to a} \\frac{f(x)}{g(x)}$ exists and $\\lim_{x \\to a} \\frac{f'(x)}{g'(x)}$ also exists, then $\\lim_{x \\to a} \\frac{f(x)}{g(x)} = \\lim_{x \\to a} \\frac{f'(x)}{g'(x)}$.<\/p>\n
In this case, we have $\\lim_{x \\to 0} \\frac{{{\\text{x}}\\sin {\\text{x}}}}{{1 – \\cos {\\text{x}}}} = \\lim_{x \\to 0} \\frac{{\\text{x}} \\cos {\\text{x}} + \\sin {\\text{x}}}{-\\sin {\\text{x}}} = \\lim_{x \\to 0} \\frac{{\\text{x}} \\cos {\\text{x}} + \\sin {\\text{x}}}{-\\sin {\\text{x}}} = \\frac{1}{2}$.<\/p>\n
Therefore, the value of $\\lim_{x \\to 0} \\frac{{{\\text{x}}\\sin {\\text{x}}}}{{1 – \\cos {\\text{x}}}}$ is $\\boxed{\\frac{1}{2}}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”1″ option2=”0″ option3=”\\[\\frac{1}{2}\\]” option4=”2″ correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20222","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n