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Limits at infinity: part 2

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Start with Limits at infinity: part 1.

Specific Limits: $$\lim_{x \to +\infty} \frac{1}{x^{r}} = 0, \quad r > 0.$$

$$\lim_{x \to -\infty} e^{x} = 0.$$

$$\lim_{x \to +\infty} \tan^{-1} x = +\frac{\pi}{2} \lim_{x \to -\infty} \tan^{-1} x = -\frac{\pi}{2}.$$


These are related to $\lim\limits_{x \to a}$ in the following sense:

Liminvtan.png

Rules of Limits. Assume $$\begin{aligned} \lim_{x \to +\infty} f(x) &= L \\ \lim_{x \to -\infty} f(x) &= M \end{aligned}$$ exist. Then $$\begin{aligned} \text{SR: } & \lim_{x \to +\infty} \left( f(x) + g(x) \right) = L + M \\ \text{DR: } & \lim_{x \to +\infty} \left( f(x) - g(x) \right) = L - M \\ \text{PR: } & \lim_{x \to +\infty} \left( f(x) \cdot g(x) \right) = LM \\ \text{QR: } & \lim_{x \to +\infty} \left( \frac{f(x)}{g(x)} \right) = \frac{L}{M}, \quad M \neq 0 \end{aligned}$$

One more with a special algebraic operation...

Substitution Rule. $$\lim_{x \to \infty} f(g(x)) = f(\lim_{x \to \infty} g(x)) $$ if $f$ is continuous at $y = lim_{x \to \infty} g(x)$.

Example. Compute $$\begin{aligned} \lim_{x \to -\infty} \left( e^{x} - \frac{1}{x} + 3 \right) &\overset{\text{SR}}{=} \lim_{x \to -\infty} e^{x} - \lim_{x \to -\infty} \frac{1}{x} + \lim_{x \to -\infty} 3 \\ & = 0 - 0 + 3 \\ & = 3 \end{aligned}$$

Example. Compute $$\lim_{x \to \infty} \frac{x}{x + 1} $$ plugging in $x = \infty$, gives us $\frac{\infty}{\infty}$ (indeterminate expression).

We are supposed to do algebra instead. But what?

Idea: Divide by $x$.

$$\begin{aligned} \lim_{x \to \infty} \frac{x}{x + 1} & = \lim_{x \to \infty} \frac{\frac{x}{x}}{\frac{(x+1)}{x}} \\ &= \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} \\ &\overset{\text{QR}}{=} \frac{\lim_{x \to \infty} 1}{\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)} \\ &\overset{\text{SR}}{=} \frac{1}{1 + 0} = 1 \end{aligned}$$

Example. Evaluate the limit at $\infty$. $$\frac{x^{2}}{x^{2} + 1}$$ Try $\frac{\infty}{\infty}$. Indeterminate, so dead end.

Dead end.png

Divide by $x$. $$\begin{aligned} \frac{x^{2}}{x^{2} + 1} & = \frac{\frac{x^{2}}{x}}{\frac{(x^{2} + 1)}{x}} \\ & = \frac{x}{x + \frac{1}{x}} \to \frac{\infty}{\infty} \end{aligned}$$ Indeterminate, still not resolved!

What to do?

Divide by $x$ again, $$\begin{aligned} \frac{x}{x + \frac{1}{x}} & = \frac{\frac{x}{x}}{\frac{(x + \frac{1}{x})}{x}} \\ &= \frac{1}{1 + \frac{1}{x^{2}}} \\ &= \frac{1}{1 + 0} \\ &= 1 \end{aligned}$$

Now $ \frac{1}{1 + \frac{1}{x^{2}}} $ doesn't give $\frac{0}{0}$ anymore!

Done, because as $x \to \infty$ $$\frac{1}{x^{2}} \to 0 $$

Better idea: Divide by $x^{2}$ in the first place.

Analysis. To evaluate the limit at $\infty$. For a rational function the leading terms of the polynomials determine the long term behavior of the numerator and denominator. $$\lim_{x \to \infty} \frac{\overbrace{3x^{3}}^{\text{Long Term}} - \overbrace{2x^{2} + x - 8}^{\text{Where}}}{\underbrace{2x^{2}}_{\text{Parabola}} - \underbrace{17x + 5}_{\text{Where it is.}}} \to \frac{\infty}{\infty} $$ What determines long term? $\frac{3x^{2}}{2x^{2}} = \frac{3}{2}x \to \infty$.

Divide numerator and denominator by $x^{2}$. $$\begin{aligned} \lim_{x \to \infty} \frac{\frac{(3x^{3} - 2x^{2} + x - 8)}{x^{2}}}{\frac{(2x^{2} - 17x + 5)}{x^{2}}} & = \lim_{x \to \infty} \frac{3x - 2 + \frac{1}{x} - \frac{8}{x^{2}}}{2 - \frac{17}{x} + \frac{5}{x^{2}}} \\ & = \frac{\infty - 2 + 0 - 0}{2 - 0 + 0} \\ & = \infty \end{aligned}$$

Note: $\frac{\infty}{\infty}$ is indeterminate, meaningless. And so is $\infty - \infty$.

Some algebra possible though: $$\begin{aligned} \infty + \infty & = \infty \\ \infty \cdot \infty &= \infty \end{aligned}$$

Example $$\lim_{x\to \infty} \left(\sqrt{x^{2} + 1} - x \right) $$ gives $\infty - \infty$. Also indeterminate. Dead end!

Dead end.png

Trick: Multiply and divide by the conjugate: $$\sqrt{x^{2} + 1} + x$$ This is the result: $$\begin{aligned} \lim_{x\to \infty} \frac{\left(\sqrt{x^{2} + 1} - x \right)\left(\sqrt{x^{2} + 1} + x\right)}{\sqrt{x^{2} + 1} + x} & = \lim_{x\to \infty} \frac{(x^{2} + 1) - x^{2}}{\sqrt{x^{2} + 1} + x} \\ & = \lim_{x \to \infty} \frac{1}{\sqrt{x^{2} + 1} + x} \\ & = \frac{1}{\infty} \\ & = 0 \end{aligned}$$

Exercise. Given $$ f(x) = \frac{x^{3} - x}{x^{2} - 6x + 5}. $$ At $\infty$, $$\begin{aligned} \lim_{x \to \infty} \frac{x^{3} - x}{x^{2} - 6x + 5} &= \lim_{x \to \infty} \frac{\frac{x^{3} - x}{x^{2}}}{\frac{x^{2} - 6x + 5}{x^{2}}} \\ & = \lim_{x \to \infty} \frac{x - \frac{1}{x}}{1 - \frac{6}{x} + \frac{5}{x^{2}}} \\ & = \frac{\infty - 0}{1 - 0 - 0} \\ & = \infty \end{aligned}$$ It's infinite, yes, but which infinity? Need to find the sign. All we need is to find the signs of the leading terms of the numerator and denominator. $$\begin{gathered} \lim_{x \to +\infty} (x^{3} - x) = +\infty, \quad \lim_{x \to -\infty} (x^{3} - x) -\infty\\ \lim_{x \to +\infty} (x^{2} - 6x + 5) = +\infty, \quad \lim_{x \to -\infty} (x^{2} - 6x + 5) = +\infty \end{gathered}$$ The $x^{3}$ and $x^{2}$ terms determines the behavior at $\infty$. Hence \begin{alignedat}{2} f+,& & \quad f-,& \\ \lim_{x \to \infty} f(x) &= +\infty, & \quad \lim_{x \to -\infty} f(x) &= -\infty \end{alignedat}

  • Horizontal Asymptotes? : None.
  • Vertical Asymptotes? : $$\frac{x^{3} - x}{x^{2} - 6x + 5}$$

Look for 0 in the denominator.

Let's factor both: $$\begin{gathered} x^{3} - x = x(x^{2} - 1) = x(x - 1)(x + 1)\\ x^{2} - 6x + 5 = (x-1)(x-5) \end{gathered}$$ Can guess or use the quadratic formula (better) $$\begin{aligned} x & = \frac{6 \pm \sqrt{6^{2} - 4\cdot 5}}{2} = \frac{6 \pm \sqrt{36 \cdot 20}}{2} \\ & = \frac{6 \pm 4}{2} = 5;1 \to (x-5)(x-1) \end{aligned}$$ Now $$f(x) = \frac{x(x-1)(x+1)}{(x-5)(x-1)} = \frac{x(x+1)}{x-5}$$ for $x \neq 1$ (Domain) $$\lim_{x \to 1} f(x) = \frac{1(1+1)}{1-5} = -\frac{1}{2}$$ because $\frac{x(x+1)}{x-5}$ is continuous at 1. So $x=1$ is not a vertical asymptote. $$\lim_{x \to 5^{-}} \frac{x(x+1)}{x-5} = \frac{30}{0} = \overset{?}{\pm} \infty$$ But which one $(\pm\infty)$? Find the sign. $$\frac{ + \; +}{-} = \; -$$ So $$\lim_{x \to 5^{-}} \frac{x(x+1)}{x-5} = - \infty$$ Similarly, $$\lim_{x \to 5^{+}} f(x) = +\infty$$ So $x=5$ is a vertical asymptote.