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Calculus

Limits

Limit properties, basic evaluations at infinity, evaluation techniques, L'Hospital's rule, and one-sided limits.

A limit is the value a function approaches as the input approaches a target, finite or ±∞. Limits formalise “arbitrarily close” and underlie continuity, derivatives, and integrals. The algebraic properties below cover routine direct substitution. The asymptotic facts cover behaviour at infinity. The evaluation techniques (factoring, conjugates, L’Hospital’s rule) handle indeterminate forms. One-sided limits cover piecewise definitions and discontinuities.

cLxyf(x)
lim x→c f(x) = L when f approaches L from both sides as x nears c.

Properties

Limits respect algebraic operations when the individual limits exist (denominator non-zero for quotients).

Constant multiple. Constants pass through:

lim⁡x→a[cf(x)]=clim⁡x→af(x)\lim_{x \rightarrow a}\left[cf\left(x\right)\right]=c\lim_{x \rightarrow a}f\left(x\right)

Sum and difference distribute:

lim⁡x→a[f(x)±g(x)]=lim⁡x→af(x)±lim⁡x→ag(x)\lim_{x \rightarrow a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim_{x \rightarrow a}f\left(x\right)\pm\lim_{x \rightarrow a}g\left(x\right)

Product of limits:

lim⁡x→a[f(x)g(x)]=lim⁡x→af(x)lim⁡x→ag(x)\lim_{x \rightarrow a}\left[f\left(x\right) g\left(x\right)\right]=\lim_{x \rightarrow a}f\left(x\right) \lim_{x \rightarrow a}g\left(x\right)

Quotient, denominator limit non-zero:

lim⁡x→a[f(x)g(x)]=lim⁡x→af(x)lim⁡x→ag(x)   provided lim⁡x→ag(x)≠0\lim_{x \rightarrow a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim_{x \rightarrow a}f\left(x\right)}{\lim_{x \rightarrow a}g\left(x\right)}\text{ }\text{ }\text{ }provided\text{ }\lim_{x \rightarrow a}g\left(x\right)\neq0

Power:

lim⁡x→a[f(x)]2=[lim⁡x→af(x)]2\lim_{x \rightarrow a}\left[f\left(x\right)\right]^{2}=\left[\lim_{x \rightarrow a}f\left(x\right)\right]^{2}

n-th root commutes with the limit:

lim⁡x→a[f(x)n]=[lim⁡x→af(x)]n\lim_{x \rightarrow a}\left[\sqrt[n]{f\left(x\right)}\right]=\sqrt[n]{\left[\lim_{x \rightarrow a}f\left(x\right)\right]}

Basic Limit Evalution at ±∞

Asymptotic facts for ±∞.

Exponential at the extremes:

lim⁡x→∞ex=∞ and lim⁡x→−∞ex=0\lim_{x \rightarrow \infty}e^{x}=\infty\text{ and }\lim_{x \rightarrow-\infty}e^{x}=0

Logarithm grows without bound and diverges to -∞ at zero:

lim⁡x→∞ln⁡x=∞ and lim⁡x→0+ln⁡x=−∞\lim_{x \rightarrow \infty}\ln^{x}=\infty\text{ and }\lim_{x \rightarrow 0^{+}}\ln^{x}=-\infty

Reciprocal powers vanish at infinity:

if r>0 then lim⁡a→∞bxr=0if\text{ }r>0\text{ }then\text{ }\lim_{a \rightarrow \infty}\frac{b}{x^{r}}=0

Extended to negative x when xr is real-valued:

if r>0 and xr is real for negative x then lim⁡a→∞bxr=0if\text{ }r>0\text{ }and\text{ }x^{r}\text{ }is\text{ }real\text{ }for\text{ }negative\text{ }x\text{ }then\text{ }\lim_{a \rightarrow \infty}\frac{b}{x^{r}}=0

Even powers go to +∞ in both directions:

n even:lim⁡a→±∞xn=∞n\text{ }even:\lim_{a\rightarrow\pm\infty}x^{n}=\infty

Odd powers keep their sign:

n odd:lim⁡a→∞xn=∞ and lim⁡a→−∞xn=−∞n\text{ }odd:\lim_{a\rightarrow\infty}x^{n}=\infty\text{ and }\lim_{a\rightarrow-\infty}x^{n}=-\infty

Even-degree polynomial:

n even:lim⁡a→±∞axn+...+bx+c=sgn(a)∞n\text{ }even:\lim_{a\rightarrow\pm\infty}ax^{n}+...+bx+c=sgn\left(a\right)\infty

Odd-degree polynomial at +∞:

n odd:lim⁡a→∞axn+...+bx+c=sgn(a)∞n\text{ }odd:\lim_{a\rightarrow\infty}ax^{n}+...+bx+c=sgn\left(a\right)\infty

Odd-degree polynomial at -∞:

n odd:lim⁡a→−∞axn+...+xx+d=sgn(a)∞n\text{ }odd:\lim_{a\rightarrow-\infty}ax^{n}+...+xx+d=sgn\left(a\right)\infty

Evalution Techniques

Tactics for evaluating limits, ordered from simplest to most general.

Continuous Function

If f is continuous at a, the limit is f(a):

if f(x) is continuous at a then lim⁡x→af(x)=f(a)if\text{ }f\left(x\right)\text{ }is\text{ }continuous\text{ }at\text{ }a\text{ }then\text{ }\lim_{x \rightarrow a}f\left(x\right)=f\left(a\right)

Continuous Functions and Composition

If the outer function is continuous at the inner function’s limit, push the limit through:

if f(x) is continuous at b and lim⁡x→ag(x)=b(a)thenif\text{ }f\left(x\right)\text{ }is\text{ }continuous\text{ }at\text{ }b\text{ }and\text{ }\lim_{x \rightarrow a}g\left(x\right)=b\left(a\right) then
lim⁡x→af(g(x))=f(lim⁡x→ag(x))=f(b)\lim_{x \rightarrow a}f\left(g\left(x\right)\right)=f\left(\lim_{x \rightarrow a}g\left(x\right)\right)=f\left(b\right)

Factor and Cancel

When direct substitution gives 0/0, factor and cancel:

lim⁡x→2x2+4x−12x2−2x=lim⁡x→2(x−2)(x+6)x(x−2)\lim_{x \rightarrow 2}\frac{x^{2}+4x-12}{x^{2}-2x}=\lim_{x \rightarrow 2}\frac{\left(x-2\right)\left(x+6\right)}{x\left(x-2\right)}

After cancellation, substitute:

=lim⁡x→2x+6x=82=4=\lim_{x \rightarrow 2}\frac{x+6}{x}=\frac{8}{2}=4

Rationalize Numberator/Denominator

For radicals, multiply by the conjugate to turn 0/0 into a cancellable form:

lim⁡x→93−xx2−81=lim⁡x→93−xx2−813+x3+x\lim_{x \rightarrow 9}\frac{3-\sqrt{x}}{x^{2}-81}=\lim_{x \rightarrow 9}\frac{3-\sqrt{x}}{x^{2}-81}\frac{3+\sqrt{x}}{3+\sqrt{x}}

Expand and cancel:

=lim⁡x→99−x(x2−81)(3+x)=lim⁡x→9−1(x−9)(3+x)=\lim_{x \rightarrow 9}\frac{9-x}{\left(x^{2}-81\right)\left(3+\sqrt{x}\right)}=\lim_{x \rightarrow 9}\frac{-1}{\left(x-9\right)\left(3+\sqrt{x}\right)}

Substitute:

=−1(18)(16)=−1108=\frac{-1}{\left(18\right)\left(16\right)}=-\frac{1}{108}

Combine Ratinal Expressions

Combine nested fractions over a common denominator to resolve 0/0:

lim⁡h→01h(1x+h−1x)=lim⁡h→01h(x−(x+h)x(x+h))\lim_{h \rightarrow 0}\frac{1}{h}\left(\frac{1}{x+h}-\frac{1}{x}\right)=\lim_{h \rightarrow 0}\frac{1}{h}\left(\frac{x-\left(x+h\right)}{x\left(x+h\right)}\right)

Simplify and cancel h:

=lim⁡h→01h(−hx(x+h))=lim⁡h→0−1x(x+h)=1x2=\lim_{h \rightarrow 0}\frac{1}{h}\left(\frac{-h}{x\left(x+h\right)}\right)=\lim_{h \rightarrow 0}\frac{-1}{x\left(x+h\right)}=\frac{1}{x^{2}}

L’Hospital’s Rule

L’Hospital’s rule applies to 0/0 and ±∞/±∞. Replace the quotient of functions by the quotient of derivatives. Reapply if the new quotient is still indeterminate. Other forms (0·∞, ∞-∞, 1∞, 00, ∞0) can be rewritten algebraically to fit. If the derivative quotient has no limit, the rule fails and the original limit must be evaluated another way.

if lim⁡x→af(x)g(x)=00 or lim⁡x→ax(x)g(x)=±∞±∞ then,if\text{ }\lim_{x \rightarrow a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}\text{ }or\text{ }\lim_{x \rightarrow a}\frac{x\left(x\right)}{g\left(x\right)}=\frac{\pm\infty}{\pm\infty}\text{ }then,

The replacement holds whether the target is finite or infinite:

lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x) a is a number, ∞ or −∞\lim_{x \rightarrow a}\frac{f\left(x\right)}{g\left(x\right)}=\lim_{x \rightarrow a}\frac{f'\left(x\right)}{g'\left(x\right)}\text{ }a\text{ }is\text{ }a\text{ }number,\text{ }\infty\text{ }or\text{ }-\infty

Polynomials at Infinity

For rational functions at infinity, factor the largest power out of numerator and denominator before evaluating.

p(x) and q(x) are polynomials. TO compute

 lim⁡x→±∞p(x)q(x) \text{ }\lim_{x \rightarrow \pm\infty}\frac{p\left(x\right)}{q\left(x\right)}\text{ }

factor largest power of x q(x) out of both p(x) and q(x) then compute limit.

Low-order terms vanish and the limit reduces to a ratio of leading coefficients (with a sign factor):

lim⁡x→−∞(3x2−45x−2x2)=lim⁡x→−∞(3−4x2x2(5x−2))=lim⁡x→−∞(3−4x25x−2)=−32\lim_{x \rightarrow-\infty}\left(\frac{3x^{2}-4}{5x-2x^{2}}\right)=\lim_{x \rightarrow-\infty}\left(\frac{3-\frac{4}{x^{2}}}{x^{2}\left(\frac{5}{x}-2\right)}\right)=\lim_{x \rightarrow-\infty}\left(\frac{3-\frac{4}{x^{2}}}{\frac{5}{x}-2}\right)=-\frac{3}{2}

Piecewise Function

The two-sided limit exists iff the one-sided limits agree.

A piecewise function split at x = -2:

lim⁡x→−2g(x) where g(x)={x2+5if x<−21−3xif x≥−2\lim_{x \rightarrow-2}g\left(x\right)\text{ }where\text{ }g\left(x\right)=\begin{cases}x^{2}+5 & if\text{ }x < -2\\ 1-3x & if\text{ }x\geq-2\end{cases}

Compute two one side limits,

Left-hand limit, x < -2 branch:

lim⁡x→−2−g(x)=lim⁡x→−2−(x2+5)=9\lim_{x \rightarrow -2^{-}}g\left(x\right)=\lim_{x \rightarrow -2^{-}}\left(x^{2}+5\right)=9

Right-hand limit, x ≥ -2 branch:

lim⁡x→−2+g(x)=lim⁡x→−2+(1−3x)=7\lim_{x \rightarrow -2^{+}}g\left(x\right)=\lim_{x \rightarrow -2^{+}}\left(1-3x\right)=7

One side limts are differnt so

The one-sided limits differ, so the two-sided limit does not exist:

 lim⁡x→−2g(x) \text{ }\lim_{x \rightarrow -2}g\left(x\right)\text{ }

doesn’t exist. if the two one sided limits had beet equal then

 lim⁡x→−2g(x) \text{ }\lim_{x \rightarrow -2}g\left(x\right)\text{ }

would have existed and had the same value.