Limits at infinity and continuity

Last week $x$ approached a number. This week there are two new questions: what does the function do as $x$ goes to infinity, and can a function be drawn “without a break”? The first brings asymptote questions, and the second brings “what should $k$ be for the function to be continuous” questions. These are almost guaranteed on the midterm. Keep in mind:

1. For a rational function, the limit at infinity is found by looking at the degrees. Divide by the highest power.
2. $\sqrt{x^2}=|x|$. As $x\to-\infty$ this means $-x$.
3. If $\infty-\infty$ involves roots, multiply by the conjugate.
4. Continuity = the limit equals the value of the function. For a piecewise function, check the junction points.
5. The Intermediate Value Theorem shows that a root exists, it does not find the root itself.

Limits at infinity

$\displaystyle\lim_{x\to\infty}f(x)=L$: as $x$ grows, $f(x)$ approaches $L$. In this case $y=L$ is a horizontal asymptote. $x\to-\infty$ is examined separately; the two directions may give different results.

Basic fact: for $p>0$, $\displaystyle\lim_{x\to\pm\infty}\frac{1}{x^p}=0$.

Rational functions: the degree rule

Divide the numerator and denominator by the highest power in the denominator.

Example 1.

$\lim_{x\to\infty}\frac{4x^3-x}{2x^3+5x^2}=\lim_{x\to\infty}\frac{4-\frac{1}{x^2}}{2+\frac{5}{x}}=\frac{4}{2}=2$

Short rule table:

If the numerator’s degree is exactly one more, there is an oblique asymptote: do polynomial division, and the quotient part is the asymptote line.

Roots and the sign trap

$\sqrt{x^2}=|x|$. That is why a minus sign appears when you pull $x$ out of a root as $x\to-\infty$.

Example 2. $f(x)=\dfrac{\sqrt{x^2+1}}{2x+3}$.

x -> +inf:  sqrt(x^2+1) = x sqrt(1 + 1/x^2)
            f = sqrt(1+1/x^2) / (2 + 3/x)  ->  1/2
x -> -inf:  sqrt(x^2+1) = |x| sqrt(1+1/x^2) = -x sqrt(1+1/x^2)
            f = -sqrt(1+1/x^2) / (2 + 3/x)  ->  -1/2

Two horizontal asymptotes: $y=\tfrac12$ and $y=-\tfrac12$.

PITFALL · pulling $x$ out of a root when $x\to-\infty$. If you write $\sqrt{x^2}=x$, the sign comes out wrong. At negative infinity, $\sqrt{x^2}=-x$.

The $\infty-\infty$ indeterminate form

Example 3. $\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)$.

Multiply and divide by the conjugate:

$\frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2}{\sqrt{1+\frac1x}+\sqrt{1-\frac1x}}\to\frac{2}{2}=1$

Exponential and inverse trigonometric functions

$\lim_{x\to\infty}e^{-x}=0,\qquad \lim_{x\to-\infty}e^{x}=0,\qquad \lim_{x\to\pm\infty}\arctan x=\pm\frac{\pi}{2}$

In ratios containing $e^x$, for $x\to\infty$ divide by $e^x$; for $x\to-\infty$ take $e^x$ directly as $0$.

Continuity

$f$ is continuous at $x=a$ if and only if:

1. $f(a)$ is defined,
2. $\displaystyle\lim_{x\to a}f(x)$ exists,
3. $\displaystyle\lim_{x\to a}f(x)=f(a)$.

Polynomials, $e^x$, $\sin x$, $\cos x$ are continuous everywhere. Rational functions, $\ln x$, $\sqrt{x}$, $\tan x$ are continuous on their domains. Sums, products and compositions of continuous functions are also continuous.

Types of discontinuity

Example 4. $f(x)=\dfrac{x^2-4}{x-2}$ ($x\ne2$), $f(2)=3$. $\lim_{x\to2}f(x)=\lim(x+2)=4\ne3=f(2)$. A removable discontinuity. If you redefine $f(2)=4$, it becomes continuous.

Example 4b (jump). $g(x)=\begin{cases}x+1, & x<0\\ x^2-2, & x\ge0\end{cases}$. At $x=0$ the left limit is $1$ and the right limit is $-2$. The limit does not exist; this is a jump discontinuity. No value of $g(0)$ can fix it; that is the difference from a removable discontinuity.

One-sided continuity and continuity on an interval

If $\lim_{x\to a^+}f(x)=f(a)$, then $f$ is continuous from the right at $a$; similarly from the left. Continuous on a closed interval $[a,b]$ means: continuous at every interior point, continuous from the right at $a$, and from the left at $b$. For example $\sqrt{x}$ is continuous on $[0,\infty)$; at $0$ only the right side is considered, because it is not defined on the left.

Composition and limits

If $g$ is continuous at $L$ and $\lim_{x\to a}f(x)=L$, then

$\lim_{x\to a}g\big(f(x)\big)=g\Big(\lim_{x\to a}f(x)\Big)=g(L)$

So a continuous function can “go inside” the limit. Example: $\displaystyle\lim_{x\to\infty}\cos\left(\frac{\pi x}{2x+1}\right)=\cos\left(\lim_{x\to\infty}\frac{\pi x}{2x+1}\right)=\cos\frac{\pi}{2}=0$. First we found the inner limit with the degree rule, then applied the cosine.

“What should k be?” questions

In a piecewise function, each piece is continuous on its own interval. The only suspicious place is the junction point. There, write value of the left piece = value of the right piece.

Example 5. $f(x)=\begin{cases} x^2+k, & x<2\\ kx-1, & x\ge2\end{cases}$

$\lim_{x\to2^-}f=4+k,\qquad f(2)=\lim_{x\to2^+}f=2k-1$

$4+k=2k-1 \Rightarrow k=5$.

If there are two unknowns, there are two junction points and you set up two equations.

The Intermediate Value Theorem

If $f$ is continuous on the closed interval $[a,b]$ and $N$ is a number between $f(a)$ and $f(b)$, there is at least one $c\in(a,b)$ with $f(c)=N$.

The special case used to find roots: if $f(a)$ and $f(b)$ have opposite signs, there is a root in between.

Example 6. Does the equation $x^3-2x-5=0$ have a root in $(2,3)$?

$f(2)=8-4-5=-1<0$, $f(3)=27-6-5=16>0$. The polynomial is continuous, so by the IVT yes. To find the root approximately, halve the interval (bisection):

f(2.5)  =  5.625 > 0   ->  the root is in (2, 2.5)
f(2.25) =  1.891 > 0   ->  the root is in (2, 2.25)
...                     ->  root ≈ 2.0946

Template to write on an exam: ”$f$ is continuous on $[a,b]$ (polynomial), $f(a)<0<f(b)$, so by the IVT there exists $c\in(a,b)$ with $f(c)=0$.” If you forget to state continuity, you lose points.

PITFALL · the IVT for a discontinuous function. For $f(x)=\tfrac1x$, $f(-1)=-1<0<1=f(1)$ but there is no root in $(-1,1)$. The function is not continuous at $x=0$, so the theorem cannot be applied.

The Max-Min Theorem

If $f$ is continuous on the closed and bounded interval $[a,b]$, it attains an absolute maximum and an absolute minimum on that interval. This theorem is the basis of the absolute extremum questions in week 8.

Common mistakes