D7 - Limits from power series

D7 - Limits from power series

A quotient can approach a definite number even when direct substitution gives zero divided by zero. You will use Maclaurin series to expose the first surviving powers, distinguish finite and infinite behaviour, and decide how many terms and which domain a limit needs.

Cancel lowest-order terms

How can (ex−1)/x(\mathrm{e}^x-1)/x approach a number when substituting zero gives 0/00/0? A limit describes values as xx approaches zero, not necessarily a value at zero itself.

Limit near zero

lim⁡x→0F(x)=L\lim_{x\to0}F(x)=L means that F(x)F(x) approaches LL as xx approaches zero through its domain. A two-sided limit requires the same behaviour from the left and right. We can cancel a factor xx while studying a limit at zero because nearby values have x≠0x\ne0.

The form 0/00/0 is indeterminate: it does not tell us the answer. For example, x/xx/x approaches 1, x2/xx^2/x approaches zero, and x/x2x/x^2 has different infinite behaviour on the two sides. The relative rates of approach matter.

Worked example 1: the constant terms cancel

Question. Find lim⁡x→0ex−1x\displaystyle\lim_{x\to0}\frac{\mathrm{e}^x-1}{x}.

Plan. Expand the exponential, cancel the constant terms and divide every remaining term by xx.

ex−1x=(1+x+x2/2+x3/6+⋯ )−1x=1+x2+x26+⋯x≠0\begin{aligned} \frac{\mathrm{e}^x-1}{x} &=\frac{(1+x+x^2/2+x^3/6+\cdots)-1}{x}\\ &=1+\frac{x}{2}+\frac{x^2}{6}+\cdots && x\ne0 \end{aligned}

The terms after the constant tend to zero. The exponential series is valid on both sides of zero.

Answer. The limit is 11.

Check. The quotient is the difference quotient [e0+x−e0]/x[\mathrm{e}^{0+x}-\mathrm{e}^0]/x. Its limit is the derivative of the exponential at zero, namely 1.

Remove one extra term from the numerator and divide by one extra power. The first surviving term moves from linear to quadratic.

Worked example 2: the linear terms cancel too

Question. Find lim⁡x→0ex−1−xx2\displaystyle\lim_{x\to0}\frac{\mathrm{e}^x-1-x}{x^2}.

Plan. Keep the exponential expansion at least through degree 2, since the constant and linear terms disappear.

ex−1−xx2=x2/2+x3/6+x4/24+⋯x2=12+x6+x224+⋯\begin{aligned} \frac{\mathrm{e}^x-1-x}{x^2} &=\frac{x^2/2+x^3/6+x^4/24+\cdots}{x^2}\\ &=\frac12+\frac{x}{6}+\frac{x^2}{24}+\cdots \end{aligned}

Answer. The limit is 1/21/2.

Check. Direct calculator values at x=0.01x=0.01 and x=−0.01x=-0.01 are approximately 0.5016710.501671 and 0.4983370.498337. They approach 1/21/2 from opposite sides. Numerical agreement supports the algebra; it does not replace the limiting argument.

Here the continuing higher-power terms tend to zero after division. The answer comes from the lowest surviving power, not from writing 0/0=00/0=0.

Watch out

Expand before substituting the limiting value. Cancelling ex\mathrm{e}^x against a term in a sum is invalid; cancellation applies to common factors after the numerator has been simplified.

Use the same second-order cancellation with a changed exponential input.

Finite and infinite limits

The first surviving powers also tell you whether a quotient remains finite. If the numerator begins with axpax^p and the denominator with bxqbx^q, where aa and bb are nonzero, its leading behaviour near zero is (a/b)xp−q(a/b)x^{p-q}.

If p>qp>q, the quotient tends to zero. If p=qp=q, it tends to a/ba/b. If p<qp<q, its magnitude grows without bound, and you must check signs on both sides. These conclusions require valid expansions near the limiting point and no earlier surviving terms.

Worked example 3: equal lowest powers

Question. Find lim⁡x→01−cos⁡xx2\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}.

Plan. Expand cosine and compare the quadratic numerator with the denominator.

1−cos⁡x=x22−x424+⋯1−cos⁡xx2=12−x224+⋯\begin{aligned} 1-\cos x&=\frac{x^2}{2}-\frac{x^4}{24}+\cdots\\ \frac{1-\cos x}{x^2}&=\frac12-\frac{x^2}{24}+\cdots \end{aligned}

Answer. The limit is 1/21/2.

Check. The identity 1−cos⁡x=2sin⁡2(x/2)1-\cos x=2\sin^2(x/2) makes the quotient 12[sin⁡(x/2)/(x/2)]2\tfrac12[\sin(x/2)/(x/2)]^2. Since the sine series gives sin⁡u/u→1\sin u/u\to1, this independently gives 1/21/2.

Change only the denominator from x2x^2 to x4x^4. The same numerator now shrinks more slowly than the denominator.

Worked example 4: a denominator that shrinks faster

Question. Determine lim⁡x→01−cos⁡xx4\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^4}.

Plan. Keep the same numerator expansion, divide by the new power and inspect the sign of its leading term.

1−cos⁡xx4=12x2−124+⋯\begin{aligned} \frac{1-\cos x}{x^4} &=\frac{1}{2x^2}-\frac1{24}+\cdots \end{aligned}

For either sign of nonzero xx, 1/(2x2)1/(2x^2) is positive and grows without bound as x→0x\to0. The remaining terms stay bounded near zero.

Answer. The quotient tends to +∞+\infty. It has no finite limit.

Check. At x=±0.1x=\pm0.1 the quotient is approximately 49.958349.9583; at x=±0.01x=\pm0.01, approximately 4999.95834999.9583. Both sides grow positively, as predicted.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account