Find the following limits.
\lim\limits_{x \to 3} f\left(x\right) when f(x)=\left(\dfrac{2}{7}\right) ^x
To find the limit, plug x = 3 into the function.
\lim\limits_{x \rightarrow 3} \left(\dfrac{2}{7}\right)^x = \left(\dfrac{2}{7}\right)^3 = \dfrac{2^3}{7^3} = \dfrac{8}{343}
\lim\limits_{x \to 3} \left(\dfrac{2}{7}\right) ^x=\dfrac8{343}
\lim\limits_{x \rightarrow -\infty}f\left(x\right) when f(x)=(3)6^{-x}
We have:
\(\displaystyle\lim\limits_{x\to -\infty}(-x)=\infty\) and \(\displaystyle\lim\limits_{x\to \infty}6^x=\infty\).
Thus, we get:
\(\displaystyle\lim\limits_{x\to -\infty}6^{-x}=\infty\)
and, as 3>0,
\(\displaystyle\lim\limits_{x\to -\infty}(3)6^{-x}=\infty\).
\lim\limits_{x \rightarrow -\infty} \left(3\right)6^{-x} = \infty
\lim\limits_{x \rightarrow \infty} f\left(x\right) when f(x)=2^{\frac{1}{4x}}
We have:
\lim\limits_{x\to \infty}(4x)=\infty and \lim\limits_{x\to\infty}\dfrac{1}{x}=0,
thus,
\lim\limits_{x\to \infty}\dfrac{1}{4x}=0.
As \lim\limits_{x\to 0}2^x=2^0=1,
we get:
\lim\limits_{x\to\infty}2^{\frac{1}{4x}}=1.
\lim\limits_{x \rightarrow \infty} 2^{\frac{1}{4x}} = 1
\lim\limits_{x \rightarrow 3} f\left(x\right) when f(x)=2\left(\dfrac{2}{5}\right)^x
To find the limit, plug x = 3 into the function.
\lim\limits_{x \rightarrow 3} 2 \left( \dfrac{2}{5} \right) ^x = 2 \left( \dfrac{2}{5} \right) ^3 = 2 \left( \dfrac{8}{125} \right) = \dfrac{16}{125}
\lim\limits_{x \rightarrow 3} 2 \left( \dfrac{2}{5} \right) ^x = \dfrac{16}{125}
\lim\limits_{x \rightarrow \infty} f\left(x\right) when f(x)=\dfrac{3^{2x}-1}{3^x}
First, simplify the function:
\dfrac{3^{2x} - 1}{3^x} = \dfrac{3^{2x}}{3^x} - \dfrac{1}{3^x} = 3^{2x - x} - \dfrac{1}{3^x} = 3^x - \dfrac{1}{3^x}
As 3>1, we have \lim\limits_{x\to\infty}3^x=\infty.
And, as \lim\limits_{x\to\infty}\dfrac{1}{x}=0,
we get:
\lim\limits_{x\to\infty}\dfrac{1}{3^x}=0.
By difference, we get:
\lim\limits_{x\to\infty}\left(3^x-\dfrac{1}{3^x}\right)=\infty
\lim\limits_{x \rightarrow \infty} \dfrac{3^{2x} - 1}{3^x} = \infty
\lim\limits_{x \rightarrow -\infty} f\left(x\right) when f(x)=\dfrac{3^x}{2^x}
First, recognize that the function can be re-arranged:
\dfrac{3^x}{2^x} = \left( \dfrac{3}{2} \right) ^x = e^{\ln\left( \left( \frac{3}{2} \right) ^x\right)} = e^{x\ln\left(\frac{3}{2}\right)}
As \ln\left(\dfrac{3}{2}\right)>0,
we have:
\lim\limits_{x\to -\infty}x\ln\left(\dfrac{3}{2}\right)=-\infty.
And, as \lim\limits_{x\to -\infty}e^x=0,
we get:
\lim\limits_{x\to -\infty}e^{x\ln\left(\frac{3}{2}\right)}=0
\lim\limits_{x \rightarrow -\infty} \dfrac{3^x}{2^x} = 0
\lim\limits_{x \rightarrow \infty} f\left(x\right) when f(x)=\dfrac{4^x}{5^x}
The function can be re-arranged:
\dfrac{4^x}{5^x} = \left( \dfrac{4}{5} \right) ^x = e^{\ln\left(\left( \frac{4}{5} \right) ^x\right)} = e^{x\ln\left(\frac{4}{5}\right)}
As \ln\left(\dfrac{4}{5}\right)<0,
we have:
\lim\limits_{x\to \infty}x\ln\left(\dfrac{4}{5}\right)=-\infty.
And, as \lim\limits_{x\to -\infty}e^x=0,
we get:
\lim\limits_{x\to \infty}e^{x\ln\left(\frac{4}{5}\right)}=0
\lim\limits_{x \rightarrow \infty} \dfrac{4^x}{5^x} = 0