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$\\ \sf \: lim_{x\longrightarrow-3}\frac{2x^2+9x+9}{2x^2+7x+3}$


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Answer 1

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Answer 2

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Answer:

Step-by-step explanation:

$\mathsf{let \;\;\frac{2x^2+9x+9}{2x^2+7x+3} \;\;be\: a\: function \:of \:x \:\; i.e. \;f(x) }\\ \\\mathsf{Notice\:that\: \lim_{x\to -3} f(-3) = \frac{0}{0}}$

$\\ \\ \sf{which\:is\:an \:intermediate\:form}\\\\\\\ \sf For\:such\:cases,we \:use \:L-Hopital's \:rule\:(i.e.\:differentiation)\\ \\ \implies \sf{ \lim_{x \to -3}}$

$\longrightarrow\sf \lim_{x \to -3} f(x)=\sf \lim_{x \longrightarrow -3}\frac{4x+9}{4x+7}$

$\longrightarrow\sf \: f(x)= \lim_{x \to -3}\frac{d}{dx} {\bigg(2x^2+9x+9\bigg)}{\frac{d}{dx}\bigg(2x^2+ 7x+3\bigg)}$

$\\\\\sf{\implies \lim_{x \to -3} f(x)= \lim_{x\to -3}\frac{4(-3)+9}{4(-3)+7}}\\\\\sf{\implies \lim_{x \to -3} f(x) = \frac{3}{5}}$