Question

evaluate the following ​

Answer 1

$\bold{ANSWER}$

$\rm{-4}$

$\mathbb{EXPLANATION}$

$\textit{Given\:Limit\:Is}$

$\displaystyle\lim_{x\to\:2}\frac{x^3-2x^2}{x^2-5x+6}$

$\rm{Put\:x=2\:in\:given\:limit\:we\:get\:\frac{0}{0}\:form}$

$\rm{in\:order\:to\:get\:rid\:from\:this\:form\:we\:use\:L'HOSPITALS\:RULE}$

$\underline{L'HOSPITALS\:\:RULE}$

$\rm{it\:states\:that\:Differentiate\:Denominator\:And\:Numenator\:seperetly}$

$\rm{till\:we\:come\:out\:from\:\frac{0}{0}\:form}$

$\displaystyle\lim_{x\to\:2}\frac{3x^2-4x}{2x-5}$

$\rm{Now\:put\:x=2\:we\:have}$

$\implies\:\frac{3\times\:2^2-4\times\:2}{2\times\:2-5}$

$\implies\:\frac{3\times\:4-4}{4-5}$

$\implies\:\frac{4}{-1}$

$\implies\:\rm{-4}$

$\therefore$ $\displaystyle\lim_{x\to\:2}\frac{x^3-2x^2}{x^2-5x+6}=\rm{-4}$