Find the following limit if it exists.
\lim\limits_{x \to -2}\dfrac{3+2x}{4^x}
We have:
\lim\limits_{x \rightarrow -2} \left(3 + 2x\right) = 3 + 2\left(-2\right) = -1
And:
\lim\limits_{x \rightarrow -2} 4^x = 4^{\left(-2\right)} = \dfrac{1}{16}
Therefore:
\lim\limits_{x \rightarrow -2} \dfrac{3 + 2x}{4^x} = \dfrac{-1}{\dfrac{1}{16}} = -16
\lim\limits_{x \to -2}\dfrac{3+2x}{4^x}=-16
\lim\limits_{x \rightarrow 1} \dfrac{6x - 1}{x^3 - 1}
We have:
\lim\limits_{x \rightarrow 1} \left(6x - 1\right) = 6\left(1\right) - 1 = 5
And:
\lim\limits_{x \rightarrow 1} \left(x^3 - 1\right) = \left(1\right)^3 - 1 = 0
Therefore:
\lim\limits_{x \rightarrow 1} \dfrac{6x - 1}{x^3 - 1} = \dfrac{5}{0} \rightarrow undefined.
The limit does not exist.
\lim\limits_{x \rightarrow -1} \dfrac{x^2 + 2x}{x^5 - 1}
We have:
\lim\limits_{x \rightarrow -1} \left(x^2 + 2x\right) = \left(-1\right)^2 + 2\left(-1\right) = 1 - 2 = -1
And:
\lim\limits_{x \rightarrow -1} \left(x^5 - 1\right) = \left(-1\right)^5 - 1 = -1 - 1 = -2
Therefore:
\lim\limits_{x \rightarrow -1} \dfrac{x^2 + 2x}{x^5 - 1} = \dfrac{-1}{-2} = \dfrac{1}{2}
\lim\limits_{x \rightarrow -1} \dfrac{x^2 + 2x}{x^5 - 1} = \dfrac{1}{2}
\lim\limits_{x \rightarrow 3} \dfrac{3x^2 + x + 1}{6x^2 + 2x}
We have:
\lim\limits_{x \rightarrow 3} \left(3x^2 + x + 1\right) = 3\left(3\right)^2 + \left(3\right) + 1 = 27 + 3 + 1 = 31
And:
\lim\limits_{x \rightarrow 3} \left(6x^2 + 2x\right) = 6\left(3\right)^2 + 2\left(3\right) = 54 + 6 = 60
Therefore:
\lim\limits_{x \rightarrow 3} \dfrac{3x^2 + x + 1}{6x^2 + 2x} = \dfrac{31}{60}
\lim\limits_{x \rightarrow 3} \dfrac{3x^2 + x + 1}{6x^2 + 2x} = \dfrac{31}{60}
\lim\limits_{x \rightarrow -2} \dfrac{3^x}{x^3}
We have:
\lim\limits_{x \rightarrow -2} 3^x = 3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}
And:
\lim\limits_{x \rightarrow -2} x^3 = \left(-2\right)^3 = -8
Therefore:
\lim\limits_{x \rightarrow -2} \dfrac{3^x}{x^3} = \dfrac{\dfrac{1}{9}}{-8} = \dfrac{-1}{72}
\lim\limits_{x \rightarrow -2} \dfrac{3^x}{x^3} = \dfrac{-1}{72}
\lim\limits_{x \rightarrow 4} \dfrac{2x - 1}{x^2 - 3^{x-2}}
We have:
\lim\limits_{x \rightarrow 4} \left(2x - 1\right) = 2\left(4\right) - 1 = 7
And:
\lim\limits_{x \rightarrow 4} \left(x^2 - 3^{x-2}\right) = 4^2 - 3^{4-2} = 16 - 3^2 = 16 - 9 = 7
Therefore:
\lim\limits_{x \rightarrow 4} \dfrac{2x - 1}{x^2 - 3^{x-2}} = \dfrac{7}{7} = 1
\lim\limits_{x \rightarrow 4} \dfrac{2x - 1}{x^2 - 3^{x-2}} = 1
\lim\limits_{x \rightarrow 3} \dfrac{5^x}{-2x^2 + 4x + 1}
We have:
\lim\limits_{x \rightarrow 3} 5^x = 5^3 = 125
And:
\lim\limits_{x \rightarrow 3} \left( -2x^2 + 4x + 1\right) = -2\left(3\right)^2 + 4\left(3\right) + 1 = -18 + 12 + 1 = -5
Therefore:
\lim\limits_{x \rightarrow 3} \dfrac{5^x}{-2x^2 + 4x + 1} = \dfrac{125}{-5} = -25
\lim\limits_{x \rightarrow 3} \dfrac{5^x}{-2x^2 + 4x + 1} = -25