[amp_mcq option1=”$$ – \\frac{1}{3}$$” option2=”$$\\frac{5}{{18}}$$” option3=”0″ option4=”$$\\frac{2}{5}$$” correct=”option2″]<\/p>\n
The correct answer is $\\boxed{\\text{D}}$.<\/p>\n
To find the limit, we can substitute $x=3$ into the function. However, this results in the indeterminate form $\\frac{0}{0}$. This means that the two expressions have the same value, but we cannot determine what that value is by simply substituting.<\/p>\n
To resolve this, we can use L’H\u00c3\u00b4pital’s rule. L’H\u00c3\u00b4pital’s rule states that if $$\\lim_{x\\to a} \\frac{f(x)}{g(x)} = \\frac{0}{0}$$ or $$\\lim_{x\\to a} \\frac{f(x)}{g(x)} = \\pm\\infty$$, then $$\\lim_{x\\to a} \\frac{f'(x)}{g'(x)}$$ exists and is equal to the limit of the original expression.<\/p>\n
In this case, we have $$\\begin{align} The other options are incorrect because they do not represent the value of the limit.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”$$ – \\frac{1}{3}$$” option2=”$$\\frac{5}{{18}}$$” option3=”0″ option4=”$$\\frac{2}{5}$$” correct=”option2″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20251","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n
\n\\lim_{x\\to 3} {\\text{f}}\\left( {\\text{x}} \\right) &= \\lim_{x\\to 3} \\frac{{2{{\\text{x}}^2} – 7{\\text{x}} + 3}}{{5{{\\text{x}}^2} – 12{\\text{x}} – 9}} \\
\n&= \\lim_{x\\to 3} \\frac{{2\\left( {\\text{x}}^2 – \\frac{7}{2}{\\text{x}} + \\frac{3}{2}} \\right)}}{{5\\left( {\\text{x}}^2 – \\frac{12}{5}{\\text{x}} – \\frac{9}{5}} \\right)}} \\
\n&= \\lim_{x\\to 3} \\frac{{2\\left( {\\text{x} – \\frac{3}{2}} \\right)}^2}}{{5\\left( {\\text{x} – \\frac{3}{5}} \\right)}^2}} \\
\n&= \\frac{{2\\left( 3 – \\frac{3}{2}} \\right)}^2}}{{5\\left( 3 – \\frac{3}{5}} \\right)}^2}} \\
\n&= \\frac{5}{18}
\n\\end{align<\/em>}$$ Therefore, $$\\mathop {\\lim }\\limits_{{\\text{x}} \\to 3} {\\text{f}}\\left( {\\text{x}} \\right) = \\frac{5}{{18}}.$$<\/p>\n