Limits: At Infinity and Infinite
Learning Objectives
- Limits at infinity
- Infinite Limits
Exercises
1.
\(\displaystyle \lim_{x\to -\infty} \dfrac{6x^5+2x}{3x^5+4x^2+18}=2\)
2.
\(\displaystyle \lim_{x\to \infty} \dfrac{-4x^3-12x^2+15}{16x^3+14x-17}=-\dfrac{1}{4}\)
\(\displaystyle \lim_{x\to -\infty} \dfrac{-4x^3-12x^2+15}{16x^3+14x-17}=-\dfrac{1}{4}\)
\(\displaystyle \lim_{x\to -\infty} \dfrac{-4x^3-12x^2+15}{16x^3+14x-17}=-\dfrac{1}{4}\)
3.
\(\displaystyle \lim_{x\to \infty} \dfrac{23x^3-15x^2-6x}{188x^2+155x+413}\rightarrow \infty\)
\(\displaystyle \lim_{x\to -\infty} \dfrac{23x^3-15x^2-6x}{188x^2+155x+413}\rightarrow -\infty\)
\(\displaystyle \lim_{x\to -\infty} \dfrac{23x^3-15x^2-6x}{188x^2+155x+413}\rightarrow -\infty\)
4.
\(\displaystyle \lim_{x\to \infty} \dfrac{4e^{2x}-6e^{-3x}}{e^{2x}+e^{-6x}+2e^{-8x}}=4\)
\(\displaystyle \lim_{x\to -\infty} \dfrac{4e^{2x}-6e^{-3x}}{e^{2x}+e^{-6x}+2e^{-8x}}=0\)
\(\displaystyle \lim_{x\to -\infty} \dfrac{4e^{2x}-6e^{-3x}}{e^{2x}+e^{-6x}+2e^{-8x}}=0\)
5.
Hole: \((-4,0)\)
VA: \(x=0,\) \(x=-9\)
\(\displaystyle \lim_{x\rightarrow-9^-} f(x)\rightarrow -\infty\)
\(\displaystyle \lim_{x\rightarrow-9^+} f(x)\rightarrow \infty\)
\(\displaystyle \lim_{x\rightarrow0^-} f(x)\rightarrow \infty\)
\(\displaystyle \lim_{x\rightarrow 0^+} f(x)\rightarrow -\infty\)
VA: \(x=0,\) \(x=-9\)
\(\displaystyle \lim_{x\rightarrow-9^-} f(x)\rightarrow -\infty\)
\(\displaystyle \lim_{x\rightarrow-9^+} f(x)\rightarrow \infty\)
\(\displaystyle \lim_{x\rightarrow0^-} f(x)\rightarrow \infty\)
\(\displaystyle \lim_{x\rightarrow 0^+} f(x)\rightarrow -\infty\)
6.
HA: \(x=0\)
The function approaches this horizontal asymptote, \(x=0,\) as \(x\rightarrow -\infty.\)
\(\displaystyle \lim_{x\rightarrow \infty} f(x)\rightarrow \infty\quad \) so there is no horizontal asymptote as \(x\rightarrow \infty.\)