Find the following limits if they exist.
\lim\limits_{x \to -1} 4^x+3x^2
The limit of a function that can be written as the sum of other functions is equal to the sum of the limits of each of the individual functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \pm g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \pm \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
For this problem, the limit can be re-written as:
\lim\limits_{x \to -1} 4^x + 3x^2 = \left( \lim\limits_{x \rightarrow -1} 4^x \right) + \left( \lim\limits_{x \rightarrow -1} 3x^2 \right)
These limits can be evaluated by plugging -1 in for x:
\left( \lim\limits_{x \rightarrow -1} 4^x \right) = 4^{-1} = \dfrac{1}{4} = 0.25
\left( \lim\limits_{x \rightarrow -1} 3x^2 \right) = 3\left(-1\right)^2 = 3
Summing the limits to get the limit of the overall function yields:
\lim\limits_{x \rightarrow -1} 4^x + 3x^2= 0.25 + 3
\lim\limits_{x \to -1} 4^x+3x^2=3.25
\lim\limits_{x \rightarrow 2} 4x^3 + \dfrac{2}{x}
The limit of a function that can be written as the sum of other functions is equal to the sum of the limits of each of the individual functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \pm g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \pm \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
For this problem, the limit can be re-written as:
\lim\limits_{x \rightarrow 2} 4x^{3} + \dfrac{2}{x} = \left( \lim\limits_{x \rightarrow 2} 4x^{3} \right) + \left( \lim\limits_{x \rightarrow 2} \dfrac{2}{x} \right)
These limits can be evaluated by plugging 2 in for x:
\left( \lim\limits_{x \rightarrow 2} 4x^{3} \right) = 4\left(2\right)^3 = 4\left(8\right) = 32
\left( \lim\limits_{x \rightarrow 2} \dfrac{2}{x} \right) = \dfrac{2}{2} = 1
Summing the limits to get the limit of the overall function yields:
\lim\limits_{x \rightarrow 2} 4x^{3} + \dfrac{2}{x} = 32 + 1 = 33
\lim\limits_{x \rightarrow 2} 4x^3 + \dfrac{2}{x} = 33
\lim\limits_{x \rightarrow 3} x^{-2} + x^{2}
The limit of a function that can be written as the sum of other functions is equal to the sum of the limits of each of the individual functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \pm g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \pm \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
For this problem, the limit can be re-written as:
\lim\limits_{x \rightarrow 3} x^{-2} + x^{2} = \left( \lim\limits_{x \rightarrow 3} x^{-2} \right) + \left( \lim\limits_{x \rightarrow 3} x^{2} \right)
These limits can be evaluated by plugging 3 in for x:
\left( \lim\limits_{x \rightarrow 3} x^{-2} \right) = 3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}
\left( \lim\limits_{x \rightarrow 3} x^{2} \right) = 3^2 = 9
Summing the limits to get the limit of the overall function yields:
\lim\limits_{x \rightarrow 3} x^{-2} + x^{2} = \dfrac{1}{9} + 9 = \dfrac{82}{9}
\lim\limits_{x \rightarrow 3} x^{-2} + x^{2} = \dfrac{82}{9}
\lim\limits_{x \rightarrow \infty } \dfrac{3x}{2x^2} + \dfrac{2x}{3}
The limit of a function that can be written as the sum of other functions is equal to the sum of the limits of each of the individual functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \pm g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \pm \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
For this problem, the limit can be re-written as:
\lim\limits_{x \rightarrow \infty } \dfrac{3x}{2x^2} + \dfrac{2x}{3} = \left( \lim\limits_{x \rightarrow \infty } \dfrac{3x}{2x^2} \right) + \left( \lim\limits_{x \rightarrow \infty } \dfrac{2x}{3} \right)
These limits can be evaluated by plugging \infty in for x:
\left( \lim\limits_{x \rightarrow \infty } \dfrac{3x}{2x^2} \right) = \left( \lim\limits_{x \rightarrow \infty } \dfrac{3}{2x} \right) = \dfrac{3}{2}\cdot \lim\limits_{x\to \infty}\dfrac{1}{x} = 0
\left( \lim\limits_{x \rightarrow \infty } \dfrac{2x}{3} \right) = \dfrac{2}{3}\cdot \lim\limits_{x\to\infty}x = \infty
Summing the limits to get the limit of the overall function yields:
\lim\limits_{x \rightarrow \infty } \dfrac{3x}{2x^2} + \dfrac{2x}{3} = 0 + \infty = \infty
\lim\limits_{x \rightarrow \infty } \dfrac{3x}{2x^2} + \dfrac{2x}{3} = \infty
\lim\limits_{x \rightarrow 4} \dfrac{5x}{4 - x} + 16x^2 - 4x
The limit of a function that can be written as the sum of other functions is equal to the sum of the limits of each of the individual functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \pm g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \pm \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
For this problem, the limit can be re-written as:
\lim\limits_{x \rightarrow 4} \dfrac{5x}{4 - x} + 16x^2 - 4x = \left( \lim\limits_{x \rightarrow 4} \dfrac{5x}{4 - x} \right) + \left( \lim\limits_{x \rightarrow 4} 16x^2 - 4x \right)
To evluate these limits let's try to plug 4 in for x:
\(\displaystyle\lim\limits_{x\to 4^-}(4-x)=0^+\) and \(\displaystyle\lim\limits_{x\to 4^+}(4-x)=0^-\).
Hence, with \lim\limits_{x\to 4}5x=5\cdot4=20, we get:
\(\displaystyle\lim\limits_{x\to 4^-}\dfrac{5x}{4-x}=\infty\) and \(\displaystyle\lim\limits_{x\to 4^+}\dfrac{5x}{4-x}=-\infty\).
Therefore:
\(\displaystyle\lim\limits_{x \to 4^-} \dfrac{5x}{4-x}\neq \lim\limits_{x \to 4^+} \dfrac{5x}{4-x
And:
\(\displaystyle\lim\limits_{x \to 4} \dfrac{5}{4-x is undefined.
\left( \lim\limits_{x \rightarrow 4} 16x^2 - 4x \right) = 16\left(4\right)^2 - 4\left(4\right) = 16\left(16\right) - 16 = 240
Since the limit of one of the functions is undefined, the overall limit is also undefined.
The limit doesn't exist.
\lim\limits_{x \rightarrow \infty } 3^{-x} + \dfrac{2x - 1}{4x^2}
The limit of a function that can be written as the sum of other functions is equal to the sum of the limits of each of the individual functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \pm g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \pm \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
For this problem, the limit can be re-written as:
\lim\limits_{x \rightarrow \infty } 3^{-x} + \dfrac{2x - 1}{4x^2} = \left( \lim\limits_{x \rightarrow \infty } 3^{-x} \right) + \left( \lim\limits_{x \rightarrow \infty } \dfrac{2x - 1}{4x^2} \right)
These limits can be evaluated by plugging \infty in for x:
\lim\limits_{x \rightarrow \infty }\left( 3^{-x} \right) = \lim\limits_{x\to \infty} \dfrac{1}{3^x} = \lim\limits_{X\to \infty}\dfrac{1}{X} = 0
\lim\limits_{x \rightarrow \infty }\left( \dfrac{2x - 1}{4x^2} \right) = \lim\limits_{x \rightarrow \infty } \left(\dfrac{1}{2x} - \dfrac{1}{4x^2} \right) = \dfrac{1}{2}\cdot \lim\limits_{x\to\infty} \dfrac{1}{x}- \dfrac{1}{4}\cdot\lim\limits_{x\to\infty}\dfrac{1}{x^2}= 0 - 0 = 0
Summing the limits to get the limit of the overall function yields:
\lim\limits_{x \rightarrow \infty } 3^{-x} + \dfrac{2x - 1}{4x^2} = 0 + 0 = 0
\lim\limits_{x \rightarrow \infty } 3^{-x} + \dfrac{2x - 1}{4x^2} = 0
\lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} + 4x^2
The limit of a function that can be written as the sum of other functions is equal to the sum of the limits of each of the individual functions:
\lim\limits_{x \rightarrow a} f\left(x\right) \pm g\left(x\right) = \left( \lim\limits_{x \rightarrow a} f\left(x\right) \right) \pm \left( \lim\limits_{x \rightarrow a} g\left(x\right) \right)
For this problem, the limit can be re-written as:
\lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} + 4x^2 = \left( \lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} \right) + \left( \lim\limits_{x \rightarrow -3} 4x^2 \right)
These limits can be evaluated by plugging -3 in for x:
\left( \lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} \right) = \dfrac{2\left(-3\right)}{3^\left(-3\right)} = \left(-6\right)\left(3^3\right) = \left(-6\right)\left(27\right) = -162
\left( \lim\limits_{x \rightarrow -3} 4x^2 \right) = 4\left(-3\right)^2 = 4\left(9\right) = 36
Summing the limits to get the limit of the overall function yields:
\lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} + 4x^2 = -162 + 36 = -126
\lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} + 4x^2 = -126