Find the following limits using factorization.
\lim\limits_{x \to 3}{\dfrac{4x-12}{2x^2-10x+12}}
Factor the numerator and the denominator.
The numerator is:
4x - 12 = 4\left(x - 3\right)
The number 2 in the denominator can be pulled out front:
2x^2 - 10x + 12 = 2\left(x^2 - 10 + 12\right)
The second-order polynomial \left(x^2 - 10 + 12\right) can be factored further:
\left(x^2 - 10 + 12\right) = \left(x - 3\right)\left(x - 2\right)
The overall function can be re-written as:
\dfrac{4x - 12}{2x^2 - 10x + 12} = \dfrac{4\left(x-3\right)}{2\left(x - 3\right)\left(x - 2\right)} = \dfrac{4}{2}\dfrac{\left(x-3\right)}{\left(x-3\right)}\dfrac{1}{x-2} = \left(2\right)\left(1\right)\left(\dfrac{1}{x-2}\right)
The limit is found by evaluating at x = 3 :
\lim\limits_{x \to 3} \dfrac{2}{x-2} = \dfrac{2}{3-2} = \dfrac{2}{1} = 2
\lim\limits_{x \to 3}{\dfrac{4x-12}{2x^2-10x+12}}=2
\lim\limits_{x \rightarrow 3} \dfrac{2x^2 - 4x - 6}{3x^2 - 3x - 18}
Factor the numerator and the denominator.
The number 2 in the numerator can be pulled out front:
2x^2 - 4x - 6 = 2\left(x^2 - 2x - 3\right)
The second-order polynomial \left(x^2 - 2x - 3\right) can be factored further:
\left(x^2 - 2x - 3\right) = \left(x - 3\right)\left(x + 1\right)
The 3 in the denominator can be pulled out front:
3x^2 - 3x - 18 = 3\left(x^2 - x - 6\right)
The second-order polynomial \left(x^2 - x - 6\right) can be factored further:
\left(x^2 - x - 6\right) = \left(x - 3\right)\left(x + 2\right)
The overall function can be re-written as:
\dfrac{2x^2 - 4x - 6}{3x^2 - 3x - 18} = \dfrac{2\left(x - 3\right)\left(x + 1\right)}{3\left(x - 3\right)\left(x + 2\right)} = \dfrac{2}{3}\dfrac{\left(x-3\right)}{\left(x-3\right)}\dfrac{\left(x+1\right)}{\left(x+2\right)} = \dfrac{2}{3}\dfrac{\left(x+1\right)}{\left(x+2\right)}
The limit is found by evaluating at x = 3 :
\lim\limits_{x \rightarrow 3} \dfrac{2}{3}\dfrac{\left(x+1\right)}{\left(x+2\right)} = \dfrac{2}{3}\dfrac{\left(3+1\right)}{\left(3+2\right)} = \dfrac{2}{3}\left( \dfrac{4}{5} \right) = \dfrac{8}{15}
\lim\limits_{x \rightarrow 3} \dfrac{2x^2 - 4x - 6}{3x^2 - 3x - 18} = \dfrac{8}{15}
\lim\limits_{x \rightarrow \frac{-4}{5}} \dfrac{5x+4}{x^2 + x + \dfrac{4}{25}}
Factor the denominator and numerator.
The second-order polynomial \left(x^2 + x + \dfrac{4}{25}\right) in the denominator can be factored further:
\left(x^2 + x + \dfrac{4}{25}\right) = \left(x + \dfrac{4}{5}\right)\left(x + \dfrac{1}{5}\right)
The number 5 can be pulled out in front of the numerator:
5x + 4 = 5\left(x + \dfrac{4}{5}\right)
The overall function can be re-written as:
\dfrac{5x+4}{x^2 + x + \dfrac{4}{25}} = \dfrac{5\left(x + \dfrac{4}{5}\right)}{\left(x + \dfrac{4}{5}\right)\left(x + \dfrac{1}{5}\right)} = \dfrac{\left(x + \dfrac{4}{5}\right)}{\left(x + \dfrac{4}{5}\right)}\times\dfrac{5}{x + \dfrac{1}{5}} = \dfrac{5}{x + \dfrac{1}{5}}
The limit is found by evaluating at x = \dfrac{-4}{5}:
\lim\limits_{x \rightarrow \frac{-4}{5}} \dfrac{5}{x + \dfrac{1}{5}} = \dfrac{5}{\dfrac{-4}{5} + \dfrac{1}{5}} = \dfrac{5}{\dfrac{-3}{5}} = \dfrac{-25}{3}
\lim\limits_{x \rightarrow \frac{-4}{5}} \dfrac{5x+4}{x^2 + x + \dfrac{4}{25}} = \dfrac{-25}{3}
\lim\limits_{x \rightarrow \frac{2}{5}} \dfrac{5x - 2}{5x^2 - 27x + 10}
Factor the numerator and the denominator.
The second-order polynomial \left(5x^2 - 27x + 10\right) can be factored further:
\left(5x^2 - 27x + 10\right) = \left(x - 5\right)\left(5x - 2\right)
The number 5 can be pulled out front of the numerator:
5x - 2 = 5\left(x - \dfrac{2}{5} \right)
The overall function can be re-written as:
\dfrac{5x - 2}{5x^2 - 27x + 10} = \dfrac{5\left( x - \dfrac{2}{5} \right)}{\left(x - 5\right)\left(5x - 2\right)} = \dfrac{\left(5x - 2\right)}{\left(5x - 2\right)} \times \dfrac{1}{x - 5} = \dfrac{1}{x - 5}
The limit is found by evaluating at x = \dfrac{2}{5} :
\lim\limits_{x \rightarrow \frac{2}{5}} \dfrac{1}{x - 5} = \dfrac{1}{\dfrac{2}{5} - 5} = \dfrac{1}{\dfrac{-23}{5}} = \dfrac{-5}{23}
\lim\limits_{x \rightarrow \frac{2}{5}} \dfrac{4x - 12}{2x^2 - 10x + 12} = \dfrac{-5}{23}
\lim\limits_{x \rightarrow \frac{1}{2}} \dfrac{2x^2 + 3x - 2}{2x - 1}
Factor the numerator and the denominator.
The second-order polynomial \left(2x^2 + 3x - 2\right) in the numerator can be factored further:
\left(2x^2 + 3x - 2\right) = \left(2x - 1\right)\left(x + 2\right)
The overall function can be re-written as:
\dfrac{2x^2 + 3x - 2}{2x - 1} = \dfrac{\left(2x - 1\right)\left(x + 2\right)}{2x - 1} = \left(x + 2\right)\times\dfrac{\left(2x - 1\right)}{\left(2x - 1\right)} = x + 2
The limit is found by evaluating at x = 3 :
\lim\limits_{x \rightarrow \frac{1}{2}} x + 2 = \dfrac{1}{2} + 2 = \dfrac{5}{2}
\lim\limits_{x \rightarrow \frac{1}{2}} \dfrac{2x^2 + 3x - 2}{2x - 1} = \dfrac{5}{2}
\lim\limits_{x \rightarrow 4} \dfrac{2x^2 - 32}{x^2 - x - 12}
Factor the numerator and the denominator.
The number 2 in the numerator can be pulled out front:
2x^2 - 32 = 2\left(x^2 - 16\right)
The second-order polynomial \left(x^2 - 16\right) can be factored further:
\left(x^2 - 16\right) = \left(x - 4\right)\left(x + 4\right)
The second-order polynomial \left(x^2 - x - 12\right) can be factored further:
\left(x^2 - x - 12\right) = \left(x - 4\right)\left(x + 3\right)
The overall function can be re-written as:
\dfrac{2x^2 - 32}{x^2 - x - 12} = \dfrac{2\left(x - 4\right)\left(x + 4\right)}{\left(x - 4\right)\left(x + 3\right)} = \dfrac{\left(x - 4\right)}{\left(x - 4\right)}\times\dfrac{2\left(x + 4\right)}{\left(x + 3\right)} = \dfrac{2\left(x + 4\right)}{\left(x + 3\right)}
The limit is found by evaluating at x = 4 :
\lim\limits_{x \rightarrow 4} \dfrac{2\left(x + 4\right)}{\left(x + 3\right)} = \dfrac{2\left(4 + 4\right)}{\left(4 + 3\right)} = \dfrac{16}{7}
\lim\limits_{x \rightarrow 4} \dfrac{2x^2 - 32}{x^2 - x - 12} = \dfrac{16}{7}
\lim\limits_{x \rightarrow 2} \dfrac{x^2 - 4}{3x^3 + 6x^2 - 12x - 24}
Factor the denominator.
The number 3 in the denominator can be pulled out front:
3x^3 + 6x^2 - 12x - 24 = 3\left(x^3 + 2x^2 - 4x - 8\right)
The third-order polynomial \left(x^3 + 2x^2 - 4x - 8\right) can be factored further:
\left(x^3 + 2x^2 - 4x - 8\right) = \left(x^2 - 4\right)\left(x + 2\right)
The overall function can be re-written as:
\dfrac{x^2 - 4}{3x^3 + 6x^2 - 12x - 24} = \dfrac{x^2 - 4}{3\left(x^2 - 4\right)\left(x + 2\right)} = \dfrac{\left(x^2 - 4\right)}{\left(x^2 - 4\right)} \times \dfrac{1}{3\left(x+2\right)} = \dfrac{1}{3\left(x + 2\right)}
The limit is found by evaluating at x = 2 :
\lim\limits_{x \rightarrow 2} \dfrac{1}{3\left(x + 2\right)} = \dfrac{1}{3\left(2 + 2\right)} = \dfrac{1}{12}
\lim\limits_{x \rightarrow 2} \dfrac{x^2 - 4}{3x^3 + 6x^2 - 12x - 24} = \dfrac{1}{12}