Question

Solve this queation .​

Answer 1

Explanation :

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$\tt \bigstar \green{lim_{ x \to3} \bigg( \cfrac{x {}^{2} - 5x + 6 }{x {}^{2} - 9} \bigg)} \\ \\ \\ \tt \underline{Apply \: L-hospital \: rule, } \\ \\ \\ \longrightarrow \tt \: lim_{ x \to3} \bigg( \cfrac{ \dfrac{d}{dx} (x {}^{2} - 5x + 6 )}{ \dfrac{d}{dx} (x {}^{2} - 9)} \bigg) \\ \\ \\ \longrightarrow \tt \: lim_{ x \to3} \bigg( \dfrac{2x - 5}{2x} \bigg) \\ \\ \\ \longrightarrow \tt \bigg( \dfrac{2 \times 3 - 5}{2 \times 3} \bigg) \\ \\ \\ \longrightarrow \tt \bigg( \dfrac{6 - 5}{6} \bigg) \\ \\ \\ \longrightarrow \tt \red{ \dfrac{1}{6} }$

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Hence :

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$\dag \boxed{\tt lim_{x \to3} \dfrac{x {}^{2} - 5x + 6 }{x {}^{2} - 9} = \dfrac{1}{6} }$

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____________________

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$\boxed{\begin{array}{|c|} \underline{ \tt \red{Some \: important\: limits, }} \\ \\ \\ \bull \tt lim_{x \to 0}(sinx) = 0 \\ \\ \\ \bull \tt lim_{x \to 0}(cosx) = 1 \\ \\ \\ \bull \tt lim_{x \to 0} \dfrac{sinx}{x} = 1 = lim_{x \to0} \dfrac{x}{sinx} \\ \\ \\ \bull \tt lim_{x \to 0} \dfrac{tanx}{x} = 1 = lim_{x \to0 } \dfrac{x}{tanx} \\ \\ \\ \bull \tt lim_{x \to 0} \dfrac{log(1 + x)}{x} = 1 \\ \\ \\ \bull \tt lim_{x \to 0}e {}^{x} = 1 \\ \\ \\ \bull \tt lim_{x \to 0} \dfrac{e {}^{x} - 1 }{x} = 1 \\ \\ \\ \bull \tt lim_{x \to 0} \dfrac{a {}^{x} - 1}{x} = log_{e}(a) \\ \\ \\ \bull \tt lim_{x \to 0} \bigg(1 + \dfrac{1}{x} \bigg) {}^{x} = e \end{array}}$

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Answer 2

$\begin{gathered} \large \mathfrak{\underline{\underline{ \green{Given :}}}} \\ \\ \rm\ \blue{lim_{ x \to3} \bigg( \cfrac{x {}^{2} - 5x + 6 }{x {}^{2} - 9} \bigg)} \\ \\ \large \mathfrak{ \underline{ \underline{ \orange{Solution :}}}} \\ \\ \boxed{ \purple{We \: have \: to \: apply \: L-hospital \: rule,} } \\ \\ \\ \implies\rm\:{ \color{navy}{ lim_{ x \to3} \bigg( \cfrac{ \dfrac{d}{dx} (x {}^{2} - 5x + 6 )}{ \dfrac{d}{dx} (x {}^{2} - 9)} \bigg) }}\\ \\ \\ \implies\rm \:{ \color{navy}{ lim_{ x \to3} \bigg( \dfrac{2x - 5}{2x} \bigg)}} \\ \\ \\ \implies \rm{ \color{navy}{ \bigg( \dfrac{2 \times 3 - 5}{2 \times 3} \bigg)}} \\ \\ \\ \implies \rm{ \color{navy}{ \bigg( \dfrac{6 - 5}{6} \bigg)}} \\ \\ \\ \implies \rm{\pink{{ \dfrac{1}{6} }}}\end{gathered}$

$\boxed{ \red{\therefore \sf{ \: \: \: lim_{ x \to3} \cfrac{x {}^{2} - 5x + 6 }{x {}^{2} - 9} = \dfrac{1}{6} }}}$