Question

Solve:


`\underset{x\rightarrow~3}{\lim}~(2x^2-18)/(x^2-3x)`
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Answer 1

`\\ \tt\longmapsto {\displaystyle{\lim_(x\to 3)}}(2x^2-18)/(x^2-3x)`

`\\ \tt\longmapsto {\displaystyle{\lim_(x\to 3)}}(2(x^2-9))/(x^2-3x)`

`\\ \tt\longmapsto {\displaystyle{\lim_(x\to 3)}}(2(x^2-3^2))/(x^2-3x)`

• (a+b)(a-b)=a^2-b^2

`\\ \tt\longmapsto {\displaystyle{\lim_(x\to 3)}}\frac{2(x+3)\cancel{(x-3)}}{x\cancel{(x-3)}}`

`\\ \tt\longmapsto {\displaystyle{\lim_(x\to 3)}}(2x+6)/(x)`

`\\ \tt\longmapsto (2(3)+6)/(3)`

`\\ \tt\longmapsto (6+6)/(3)`

`\\ \tt\longmapsto (12)/(3)=4`

Answer 2

Hello, please consider the following.

`\displaystyle \lim_(x\rightarrow3)~(2x^2-18)/(x^2-3x) \\ \\ \\ =\lim_(x\rightarrow3)~(2(x^2-3^2))/(x(x-3)) \\ \\ \\ =\lim_(x\rightarrow3)~(2(x-3)(x+3))/(x(x-3)) \\ \\ \\ =\lim_(x\rightarrow3)~(2(x+3))/(x) \\ \\ \\=(2(3+3))/(3)\\ \\ \\=(2*3*2)/(3) =\Large \boxed{\sf \bf \ 4 \ }`

Thank you