Introduction to Limits

The limit of $f(x)$ as $x$ approaches $a$ is the single value $L$ that the outputs $f(x)$ get arbitrarily close to as $x$ gets close to $a$ (from both sides), without necessarily equaling $a$. We write

The value $f(a)$ itself may be different from $L$, or may not even exist—the limit only cares about the behavior near $a$.

If $\lim\limits_{x\to a}f(x)$ and $\lim\limits_{x\to a}g(x)$ both exist, then limits distribute over the arithmetic:

Direct substitution: for any polynomial, and for any rational/radical function where the denominator is nonzero and the radicand is valid, $\displaystyle\lim_{x\to a}f(x)=f(a)$.

The right-hand limit $\displaystyle\lim_{x\to a^{+}}f(x)$ uses inputs slightly larger than $a$; the left-hand limit $\displaystyle\lim_{x\to a^{-}}f(x)$ uses inputs slightly smaller. The two-sided limit exists only when they agree:

A limit fails to exist (DNE) when: the one-sided limits differ (a jump), the function grows without bound (infinite), or the function oscillates wildly near $a$.

If $f(x)$ increases or decreases without bound as $x\to a$, we write $\displaystyle\lim_{x\to a}f(x)=\infty$ (or $-\infty$). This is a way of describing a limit that fails to exist; the line $x=a$ is a vertical asymptote. For $f(x)=\dfrac1x$:

$\displaystyle\lim_{x\to\infty}f(x)=L$ means the outputs level off at the height $y=L$ as $x$ grows large; the line $y=L$ is a horizontal asymptote. Two building blocks:

Rational functions $\dfrac{p(x)}{q(x)}$—compare degrees:

• degree top $<$ degree bottom $\Rightarrow$ limit $0$ (H.A. $y=0$);
• degree top $=$ degree bottom $\Rightarrow$ limit $=$ ratio of leading coefficients;
• degree top $>$ degree bottom $\Rightarrow$ limit $\pm\infty$ (no H.A.).

A function $f$ is continuous at $x=a$ if all three hold:

Informally: you can draw the graph through $x=a$ without lifting your pencil. If any part fails, $f$ has a discontinuity there.

• Removable (a hole): $\lim\limits_{x\to a}f$ exists but $\ne f(a)$ (or $f(a)$ is undefined). “Removable” because redefining one point would fix it.
• Jump: the left- and right-hand limits both exist but disagree.
• Infinite: the function blows up to $\pm\infty$ (a vertical asymptote).

The slope of the secant line through $\big(x,f(x)\big)$ and $\big(x+h,f(x+h)\big)$ is the difference quotient $\dfrac{f(x+h)-f(x)}{h}$. Letting $h\to 0$ slides the second point into the first; the secant slopes approach the slope of the tangent line. That limiting slope is the derivative:

The other great problem of calculus is finding the area under a curve $y=f(x)$ over $[a,b]$. Approximate it with $n$ thin rectangles of width $\Delta x=\tfrac{b-a}{n}$, sum their areas, and take a limit:

As $n\to\infty$ the rectangles shrink and the approximation becomes exact. This limit is the definite integral.

When direct substitution gives $\tfrac{0}{0}$, the expression hides a common factor you must expose and cancel. Choose the tool by what you see:

• Polynomial ratio $\Rightarrow$ factor top and bottom, cancel the shared factor.
• A square root $\Rightarrow$ multiply by the conjugate to turn $\sqrt{\ }$ differences into plain differences.
• A complex fraction (fractions stacked inside a fraction) $\Rightarrow$ combine the small fractions over a common denominator first, then simplify the big fraction.
• A trig ratio $\Rightarrow$ rewrite in terms of $\sin$ and $\cos$ and steer toward the small-angle facts $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ and $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$.

The floor $\lfloor x\rfloor$ is the greatest integer $\le x$; its graph is a staircase of flat steps with jumps at every integer. So at an integer $n$ the one-sided limits disagree:

Between integers the floor is constant, so for any non-integer $c$, $\displaystyle\lim_{x\to c}\lfloor x\rfloor=\lfloor c\rfloor$ (continuous there).

Suppose a piecewise $f$ has two unknown constants $a,b$ and you want it smooth (continuous and differentiable) at the split point $x=c$. You get two equations from two conditions:

Match the heights for continuity and the slopes for differentiability, then solve the resulting $2\times 2$ system for $a$ and $b$.