Find the following limits.
\lim\limits_{x \to \infty}2^{-x}.\left(1+\dfrac2x \right)
The limit of a function that can be written as the product of two other functions is equal to the product of the limit of those two functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \cdot g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \cdot \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
The function can be re-written as f\left(x\right) \cdot g\left(x\right), with:
- f\left(x\right) = 2^{-x}
- g\left(x\right) = 1+\dfrac{2}{x}
Then:
\lim\limits_{x \rightarrow \infty} 2^{-x} \cdot \left(1+\dfrac{2}{x}\right) = \lim\limits_{x \rightarrow \infty}2^{-x} \cdot \lim\limits_{x \rightarrow \infty} \left(1+\dfrac{2}{x}\right)
We have:
\lim\limits_{x \rightarrow \infty}2^{-x}=\lim\limits_{x \rightarrow \infty} \dfrac{1}{2^{x}} = \lim\limits_{t \rightarrow \infty}\dfrac{1}{t} = 0
And:
\lim\limits_{x \rightarrow \infty} \left(1+\dfrac{2}{x}\right) = 1 + 2\times \lim\limits_{x \rightarrow \infty}\dfrac{1}{x} = \left(1 + 0\right) = 1
Therefore:
\lim\limits_{x \rightarrow \infty} 2^{-x} \cdot \left(1+\dfrac{2}{x}\right)=0 \cdot 1 = 0
\lim\limits_{x \to \infty}2^{-x}.\left(1+\dfrac2x \right)=0
\lim\limits_{x \rightarrow -2} x^{3} \cdot 3x^{2}
The limit of a function that can be written as the product of two other functions is equal to the product of the limit of two others functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \cdot g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \cdot \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
The function can be re-written as f\left(x\right) \cdot g\left(x\right), with:
- f\left(x\right) = x^{3}
- g\left(x\right) = 3x^{2}
Then:
\lim\limits_{x \rightarrow -2} x^{3} \cdot 3x^{2} = \lim\limits_{x \rightarrow -2} x^{3} \cdot \lim\limits_{x \rightarrow -2} 3x^{2}
We have:
\lim\limits_{x \rightarrow -2} x^{3} = \left(-2\right)^3 = -8
And:
\lim\limits_{x \rightarrow -2} 3x^{2} = 3\left(-2\right)^2 = 3\left(4\right) = 12
Therefore:
\lim\limits_{x \rightarrow -2} x^{3} \cdot 3x^{2} = -8 \cdot 12 = -96
\lim\limits_{x \rightarrow -2} 2^{-x} \cdot \left( 1+\dfrac{2}{x} \right) = -96
\lim\limits_{x \rightarrow \infty} \dfrac{2x + 1}{4x^2} \cdot \dfrac{3x^3 + 3}{x}
The limit of a function that can be written as the product of two other functions is equal to the product of the limit of those two functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \cdot g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \cdot \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
The function can be re-written as f\left(x\right) \cdot g\left(x\right), with:
- f\left(x\right) = \dfrac{2x + 1}{4x^2}
- g\left(x\right) = \dfrac{3x^3 + 3}{x}
Then:
\lim\limits_{x \rightarrow \infty} \dfrac{2x + 1}{4x^2} \cdot \dfrac{3x^3 + 3}{x} = \lim\limits_{x \rightarrow \infty} \dfrac{2x + 1}{4x^2} \cdot \lim\limits_{x \rightarrow \infty} \dfrac{3x^3 + 3}{x}
We have:
\lim\limits_{x \rightarrow \infty} \dfrac{2x + 1}{4x^2} = \lim\limits_{x \rightarrow \infty}\left( \dfrac{2x}{4x^2} + \dfrac{1}{4x^2}\right) = \lim\limits_{x \rightarrow \infty} \dfrac{1}{2x} + \dfrac{1}{4x^2} = \dfrac{1}{2}\times \lim\limits_{x \rightarrow \infty}\dfrac{1}{x} + \dfrac{1}{4}\times \lim\limits_{x \rightarrow \infty}\dfrac{1}{x^2} = 0 + 0 = 0
And:
\lim\limits_{x \rightarrow \infty} \dfrac{3x^3 + 3}{x} = \lim\limits_{x \rightarrow \infty} \left(\dfrac{3x^3}{x} + \dfrac{3}{x}\right) = \lim\limits_{x \rightarrow \infty} \left(3x^2 + \dfrac{3}{x}\right) = 3\times \lim\limits_{x \rightarrow \infty}x^2 + 3\times \lim\limits_{x \rightarrow \infty}\dfrac{1}{x} = \infty + 0 = \infty
Therefore:
\lim\limits_{x \rightarrow \infty} \dfrac{2x + 1}{4x^2} \cdot \dfrac{3x^3 + 3}{x} = 0 \cdot \infty = 0
\lim\limits_{x \rightarrow \infty} 2^{-x} \cdot \left( 1+\dfrac{2}{x} \right) = 0
\lim\limits_{x \rightarrow 3} 2^{x} \cdot \dfrac{4}{x - 3}
The limit of a function that can be written as the product of two other functions is equal to the product of the limit of those two functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \cdot g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \cdot \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
The function can be re-written as f\left(x\right) \cdot g\left(x\right), with:
- f\left(x\right) = 2^{x}
- g\left(x\right) = \dfrac{4}{x - 3}
Then:
\lim\limits_{x \rightarrow 3} 2^{x} \cdot \dfrac{4}{x - 3} = \lim\limits_{x \rightarrow 3} 2^{x} \cdot \lim\limits_{x \rightarrow 3} \dfrac{4}{x - 3}
We have:
\lim\limits_{x \rightarrow 3} 2^{x} = 2^{3} = 8
And:
\(\displaystyle\lim\limits_{x\to 3^-}(x-3)=0^-\) and \(\displaystyle\lim\limits_{x\to 3^+}(x-3)=0^+\).
Hence:
\(\displaystyle\lim\limits_{x\to 3^-}\dfrac{4}{x-3}=-\infty\) and \(\displaystyle\lim\limits_{x\to 3^+}\dfrac{4}{x-3}=\infty\).
Therefore:
\(\displaystyle\lim\limits_{x \to 3^-} \dfrac{4}{x - 3}\neq \lim\limits_{x \to 3^+} \dfrac{4}{x - 3
And:
\(\displaystyle\lim\limits_{x \to 3} \dfrac{4}{x - 3 is undefined.
The limit does not exist.
\lim\limits_{x \rightarrow 0} 3^{-x} \cdot \dfrac{2}{3} \left(x + 3\right)^2
The limit of a function that can be written as the product of two other functions is equal to the product of the limit of those two functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \cdot g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \cdot \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
The function can be re-written as f\left(x\right) \cdot g\left(x\right), with:
- f\left(x\right) = 3^{-x}
- g\left(x\right) = \dfrac{2}{3} \left(x + 3\right)^2
Then:
\lim\limits_{x \rightarrow 0} 3^{-x} \cdot \dfrac{2}{3} \left(x + 3\right)^2 = \lim\limits_{x \rightarrow 0} 3^{-x} \cdot \lim\limits_{x \rightarrow 0} \dfrac{2}{3} \left(x + 3\right)^2
We have:
\lim\limits_{x \rightarrow 0} 3^{-x} = 3^{0} = 1
And:
\lim\limits_{x \rightarrow 0} \dfrac{2}{3} \left(x + 3\right)^2 = \dfrac{2}{3} \left(0 + 3\right)^2 = \dfrac{2}{3} \left(3^2\right) = \left(2\right)\left(3\right) = 6
Therefore:
\lim\limits_{x \rightarrow 0} 3^{-x} \cdot \dfrac{2}{3} \left(x + 3\right)^2 = 1 \cdot 6 = 6
\lim\limits_{x \rightarrow 0} 2^{-x} \cdot \left( 1+\dfrac{2}{x} \right) = 6
\lim\limits_{x \rightarrow -1} \dfrac{3x^2}{x + 2} \cdot \dfrac{1}{x}
The limit of a function that can be written as the product of two other functions is equal to the product of the limit of those two functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \cdot g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \cdot \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
The function can be re-written as f\left(x\right) \cdot g\left(x\right), with:
- f\left(x\right) = \dfrac{3x^2}{x + 2}
- g\left(x\right) = \dfrac{1}{x}
Then:
\lim\limits_{x \rightarrow -1} \dfrac{3x^2}{x + 2} \cdot \dfrac{1}{x} = \lim\limits_{x \rightarrow -1} \dfrac{3x^2}{x + 2} \cdot \lim\limits_{x \rightarrow -1} \dfrac{1}{x}
We have:
\lim\limits_{x \rightarrow -1} \dfrac{3x^2}{x + 2} = \dfrac{3\left(-1\right)^2}{-1 + 2} = \dfrac{3}{1} = 3
And:
\lim\limits_{x \rightarrow -1} \dfrac{1}{x} = \dfrac{1}{-1} = -1
Therefore:
\lim\limits_{x \rightarrow -1} \dfrac{3x^2}{x + 2} \cdot \dfrac{1}{x} = 3 \cdot -1 = -3
\lim\limits_{x \rightarrow -1} 2^{-x} \cdot \left( 1+\dfrac{2}{x} \right) = -3
\lim\limits_{x \rightarrow 5} 5^{-x} \cdot 3x^{6}
The limit of a function that can be written as the product of two other functions is equal to the product of the limit of those two functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \cdot g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \cdot \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
The function can be re-written as f\left(x\right) \cdot g\left(x\right), with:
- f\left(x\right) = 5^{-x}
- g\left(x\right) = 3x^{6}
Then:
\lim\limits_{x \rightarrow 5} 5^{-x} \cdot 3x^{6} = \lim\limits_{x \rightarrow 5} 5^{-x} \cdot \lim\limits_{x \rightarrow 5} 3x^{6}
We have:
\lim\limits_{x \rightarrow 5} 5^{-x} = 5^{-5} = \dfrac{1}{5^5}
And:
\lim\limits_{x \rightarrow 5} 3x^{6} = 3\left( 5^{6} \right)
Therefore:
\lim\limits_{x \rightarrow 5} 5^{-x} \cdot 3x^{6} = \dfrac{1}{5^5} \cdot 3 \left( 5^{6} \right) = 3 \left( \dfrac{1}{5^5} \cdot 5^{6} \right) = 3 \left( \dfrac{5^6}{5^5} \right) = 3 \left( 5^{6-5} \right) = 3 \left( 5^{1} \right) = 15
\lim\limits_{x \rightarrow 5} 2^{-x} \cdot \left( 1+\dfrac{2}{x} \right) = 15