Calc BC Unit 2 — L’Hôpital & Improper Integrals

Limits, derivative rules, L’Hôpital, improper-integral setups, p-test, and direct comparison.

First step for every limit

Plug in the target value or check the behavior at infinity, then write the form.

When is L’Hôpital allowed directly?

Only for 0/0 or infinity/infinity.

L’Hôpital rule

Differentiate numerator and denominator separately: f/g becomes f'/g'. Never use the quotient rule.

If the form is infinity minus infinity

Combine or rewrite into one fraction, then check the new form.

If the form is 0 times infinity

Rewrite as a quotient, then check the new form.

For 0^0, 1^infinity, or infinity^0

Let y equal the expression; take ln(y); find L = limit ln(y); then y = e^L.

Power rule derivative

d/dx[x^n] = n x^(n-1).

Chain rule

Derivative of outside times derivative of inside.

Derivative of 1/x

-1/x^2.

Derivative of sqrt(x)

1/(2 sqrt(x)).

Derivative of e^u

e^u times u'.

Derivative of ln(u)

u'/u.

Derivative of sin(u)

cos(u) times u'.

Derivative of cos(u)

-sin(u) times u'.

Derivative of tan(u)

sec^2(u) times u'.

Derivative of sec(u)

sec(u) tan(u) times u'.

Derivative of tan(1/x)

sec^2(1/x) times (-1/x^2); the inside derivative is required.

Derivative of ln(1 + 1/x)

(-1/x^2)/(1 + 1/x).

What does x to infinity do to 1/x?

1/x approaches 0.

Key trig values at 0

sin(0)=0; tan(0)=0; cos(0)=1; sec(0)=1.

Trig identities for tan and sec

tan(x)=sin(x)/cos(x); sec(x)=1/cos(x).

Meaning of x to 0+

x approaches 0 through positive values, e.g. 0.1, 0.01, 0.001.

What is 1/(0+)?

Positive infinity.

What is 1/(0-)?

Negative infinity.

If plugging in gives nonzero/0

Check left/right signs. It is not indeterminate, so do not use L’Hôpital.

Combine 1/x - 1/sqrt(x)

1/sqrt(x)=sqrt(x)/x, so it becomes [1-sqrt(x)]/x.

Combine 1/x - 1/(cube root x)

1/x^(1/3)=x^(2/3)/x, so it becomes [1-x^(2/3)]/x.

Antiderivative power rule

Integral x^n dx = x^(n+1)/(n+1) + C, for n not equal to -1.

Integral of 1/x

ln|x| + C.

Integral of 1/(x-a)

ln|x-a| + C.

Integral of e^(ax)

(1/a)e^(ax) + C.

Basic trig antiderivatives

Integral cos x dx=sin x+C; integral sin x dx=-cos x+C; integral sec^2 x dx=tan x+C.

Integral of 1/(x^2+a^2)

(1/a) arctan(x/a)+C.

Fundamental Theorem of Calculus

If F'=f, then integral from a to b f(x)dx = F(b)-F(a).

Integration by parts

Integral u dv = uv - integral v du.

When is an integral improper?

When it has an infinite bound, an undefined/unbounded endpoint, or a vertical asymptote inside the interval.

Set up integral from a to infinity

lim b to infinity of integral from a to b f(x)dx.

Set up integral from negative infinity to a

lim t to negative infinity of integral from t to a f(x)dx.

Bad left endpoint a

lim c to a+ of integral from c to b f(x)dx.

Bad right endpoint b

lim c to b- of integral from a to c f(x)dx.

Vertical asymptote inside an integral

Split at the asymptote and use one-sided limits on both pieces; both must converge.

Integral from negative infinity to infinity

Split at c: lim t to -infinity integral t to c plus lim b to infinity integral c to b. Both must converge.

Improper integral converges when

Every required limit exists and is finite.

Improper integral diverges when

At least one required limit is infinite or does not exist.

p-test at infinity

Integral from 1 to infinity of 1/x^p converges if p>1 and diverges if p<=1.

p-test near zero

Integral from 0 to 1 of 1/x^p converges if p<1 and diverges if p>=1.

Direct comparison: convergence

If 0<=f<=g and integral g converges, then integral f converges.

Direct comparison: divergence

If 0<=g<=f and integral g diverges, then integral f diverges.

Comparison-test memory trick

Smaller than a convergent integral means convergent; larger than a divergent integral means divergent.

What to show in a comparison answer

A correct inequality, the known p-integral, and your converges/diverges conclusion.

Why does integral 0 to 1 of 1/sqrt(x) converge?

It is 1/x^(1/2) near zero, and p=1/2<1.

Why does integral 1 to infinity of 1/sqrt(x) diverge?

It is 1/x^(1/2) at infinity, and p=1/2<=1.

Can symmetric infinite pieces cancel?

No. Split the improper integral into separate pieces; both must converge.

Can +infinity and -infinity cancel in an improper integral?

No. Each one-sided limit must be finite.

Improper-integral checklist

Find every bad point/bound, replace it with the correct limit, evaluate or compare, and state converges/diverges.