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

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Answer 1

Answer:

3/5

Step-by-step explanation:

$\mathsf{let \;\;\frac{2x^2+9x+9}{2x^2+7x+3} \;\;be\: a\: function \:of \:x \:\; i.e. \;f(x) }\\\\\mathsf{Notice\:that\: \lim_{x\to -3} f(-3) = \frac{0}{0} },\\\\\sf{which\:is\:an \:intermediate\:form}\\\\\\\ For\:such\:cases,we \:use \:L-Hopital's \:rule\:(i.e.\:differentiation)\\\\\implies \sf{ \lim_{x \to -3} f(x)= \lim_{x \to -3} \frac{\frac{d}{dx} (2x^2+9x+9)}{\frac{d}{dx}(2x^2+ 7x+3)}}}\\\\\implies\sf{ \lim_{x \to -3} f(x)= \lim_{x \to -3}\frac{4x+9}{4x+7} }$

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

Answer 2

$\tt \lim_{(x \to - 3)} \: f(x) = \frac{4( - 2) + 9}{4( - 2) + 7} = \frac{3}{5}$