Course Notes for Calculus I

I use asymptotic analysis for limits as \(x \rightarrow \infty\text{.}\) The asymptotic order of the numerator is \(x^2\text{.}\) The asymptotic order of the denominator is \(x^4\text{.}\) The denominator has a higher asymptotic order, so the limit is 0.

I use asymptotic analysis for limits as \(x \rightarrow \infty\text{.}\) The asymptotic order of the numerator is \(x^5\text{.}\) The asymptotic order of the denominator is \(x^3\text{.}\) The numerator has a higher asymptotic order and the leading coefficients are both positive, so the limit diverges to \(\infty\text{.}\)

I use asymptotic analysis for limits as \(x \rightarrow \infty\text{.}\) The asymptotic order of the numerator is determined by ignoring the 1. That leaves \(\sqrt{x^5} = x^\frac{5}{2}\text{.}\) The asymptotic order of the denominator is \(x^2\text{.}\) Since 2 is less than \(\frac{5}{2}\text{,}\) the numerator has a higher asymptotic order. Since the leading coefficients are positive and the square root produces a positive number, this limit diverges to \(\infty\text{.}\)

I use asymptotic analysis for limits as \(x \rightarrow \infty\text{.}\) The asymptotic order of the numerator is \(x\text{.}\) The asymptotic order of the denominator is found by ignoring the 1, giving \(\sqrt[3]{x^5} = x^\frac{5}{3}\text{.}\) Since \(\frac{5}{3}\) is greater than 1, the denominator has higher asymptotic order and the limit converges to 0.

I use asymptotic analysis for limits as \(x \rightarrow \infty\text{.}\) The asymptotic order of the numerator is \(x^2e^x\text{.}\) The asymptotic order of the denominator is \(e^{2x}\text{.}\) By the laws of exponents, \(e^{2x} = (e^x)^2 = e^x e^x \text{.}\) If I compare the two asymptotic orders and remove \(e^x\) form each, what is left in the numerator is \(x^2\) amd what is left in the denominator is \(e^x\text{.}\) In this way, I see that the denominator has higher asymptotic order, and the limit converges to 0.

I use asymptotic analysis for limits as \(x \rightarrow \infty\text{.}\) The asymptotic order of the numerator is \(\ln x\text{.}\) The asymptotic order of the denominator is \(\sqrt{x}\text{.}\) The denominator has higher asymptotic order, so the limit converges to 0.

I use asymptotic analysis for limits as \(x \rightarrow \infty\text{.}\) The asymptotic order of the numerator is \(e^{7x}\text{.}\) The asymptotic order of the denominator is \(e^{8x}\text{.}\) By laws of exponents, I can write \(e^{7x} = (e^x)^7\) and \(e^{8x} = (e^x)^8 = (e^x)^7(e^x)\text{.}\) In this way I see that the denominator has a higher asymptotic order, since I multiply by \(e^x\) an additional time. Therefore, the limit converges to 0.

The denominator has higher asymptotic order: \(x^2 > x\text{.}\) Therefore, the limit approaching \(\infty\) is 0, meaning that \(y=0\) is a horizontal asymptote in the positive direction. In the negative direction, the result is the same, so \(y=0\) is a horizontal asymptote in both directions.

As \(x \rightarrow \infty\text{,}\) since the asymptotic orders of numerator and denominator are the same, this approaches the ration of the leading terms, which is \(\frac{8}{17}\text{.}\) Therefore, \(y=\frac{8}{17}\) is a horizontal asymptote in the positive direction. In the negative direction, the behaviour is the same. Both of the \(x^3\) terms will be negative, but the negatives will cancel and the limit will still be \(\frac{8}{17}\text{,}\) making \(y=\frac{8}{17}\) a horizontal asymptote in both directions.

The asymptotic order of both numerator and denominator is \(\ln x\text{.}\) Since they are the same, the limit as \(x \rightarrow \infty\) is the limit of the leading coefficients: \(\frac{8}{-12} = \frac{-2}{3}\text{.}\) Therefore, \(y=\frac{-2}{3}\) is a horizontal asymptote in the positive direction. There cannot be any horizontal asymptote in the negative direction, since the domain of the function is restricted to positive number by the logarithm terms.

As \(x \rightarrow \infty\text{,}\) the denominator of the fraction \(\frac{4}{3-x}\) approaches \(-\infty\text{.}\) Since the numerator is fixed, the fraction itself approach 0. That leaves the limit as \(\frac{1}{6-0} = \frac{1}{6}\text{.}\) Therefore, \(y = \frac{1}{6}\) is a horizontal asymptote in the positive direction. The same logic works for \(x \rightarrow -\infty\text{;}\) the only difference is that the denominator of the fraction is positive, but the conclusion is the same. Therefore, \(y = \frac{1}{6}\) is a horizontal asymptote in both directions.

Unlike the previous question, this is determined not by a calculation, but by familiarity with the arctangent function. By itself, arctangent has a horizontal asymptote of \(y = \frac{\pi}{2}\) in the positive direction and \(y = \frac{-\pi}{2}\) in the negative direction. Multiplying by 8 means that there is a horizontal asymptote of \(y = 4\pi\) in the positive direction and \(y = -4\pi\) in the negative direction. Subtracting 4 means that there is a horizontal asymptote of \(y = 4\pi - 4 \) in the positive direction and \(y = -4\pi - 4 \) in the negative direction.

Since there is \(x^2\) in the function (and no other appearances of the independent variable), the behaviour in the positive and negative directions will be the same. In either direction, as \(x \rightarrow \infty\text{,}\) this is exponential decay to zero. Therefore, \(y=0\) is a horizontal asymptote in both directions.

This is an unbounded exponential function, so the limit is \(\infty\text{.}\) This is exponential growth, with a starting value of 140 and a growth rate coefficient of \(\frac{1}{40}\text{.}\)

Both numerator and denominator have the same asymptotic order, \(e^{\frac{1}{37} t}\text{,}\) so I look at the leading coefficients for the limit: \(\frac{75}{420} = \frac{5}{28}\text{.}\) The long term behaviour of this is stabilitzation at the value \(\frac{5}{28}\text{.}\) This is logistic growth, since it fits the specialized logistic growth form with the same exponential functions in the numerator and denominator.

The asymptotic order of the numerator and denominator are both \(t^2\text{,}\) so the limit as \(t \rightarrow \infty\) is the ratio of the leading coefficients, which is 3. Even though this levels off, this is not logistic growth, since it doesn’t conform to the specific construction of logistic growth with equivalent exponentials in both numerator and denominator.

Looking at the sinusoidal term, the fraction before the sine function gives an amplitude. In the limit, this fraction has the same asymptotic order in the numerator and denominator, to its limit is \(\frac{1}{9}\text{,}\) the ratio of its leading coefficient. The long term behaviour of this function is oscillation around the equilibrium value of 400 with an amplitude of \(\frac{1}{9}\text{.}\) All these values are close to 400, but the function never approaches any one value; instead, the oscillations continue for ever. This is neither exponential or logistic growth.

The exponential coefficient in the denominator is smaller than the exponential coefficient the numerator, so the asymptotic order of the denominator is lower and the long term behaviour is growth to infinity. This looks very close to logistic growth, but it isn’t: the exponential coefficients are not the same, and its doesn’t level off to a non-zero carrying capacity.

This is an exponential decay function. The long term behaviour is decay to zero.