Module 15: Newton’s method, indeterminate forms, and growth

8 exercises, each with a complete solution. 1 figure. 1 note from the engine.

On this page

A. Prerequisites and observable goals
B. The central idea: an approximation and a theorem are different tools
C. Precise tools and caution signs
D. Why Newton’s formula and the limit rule work
E. A decision procedure
F. Worked teaching examples
G. Independent practice
H. Complete solutions: Module 15
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Tangent lines, limits, chain and quotient rules, and logarithms.

After

Perform and assess Newton iterations; recognize the hypotheses for L’Hôpital’s rule; and transform other indeterminate forms without illegal cancellation.

B. The central idea: an approximation and a theorem are different tools#

Newton’s method uses tangent lines to approximate a root. It is an algorithm, not a guarantee of convergence. L’Hôpital’s rule is a theorem about certain limits of quotients. It is not the quotient rule and does not say that two quotients are equal away from a limit. Growth comparisons ask which term dominates in a limiting regime; size at one finite input is not enough.

Newton’s method follows tangents. For

f(x)=x^2-2

, start at

x_0=2

. The tangent there crosses the axis at

x_1=1.5

, and the tangent at

x_1

crosses at

x_2\approx1.4167

, already within

0.003

of the root

\sqrt2\approx1.4142

. Each step replaces the curve by its tangent and solves the easier problem.

C. Precise tools and caution signs#

Given $f$ and an initial guess $x_0$ , Newton’s iteration is

x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},

provided the denominator exists and is nonzero. A tiny derivative can cause a huge jump; cycles, divergence, or leaving the function’s domain are possible. Near a simple root, sufficient smoothness and a sufficiently close initial guess can yield rapid convergence. Do not replace those conditions by “Newton always works.”

A useful version of L’Hôpital’s rule: on a one-sided punctured interval approaching a finite endpoint or infinity, assume $f,g$ are differentiable, $g'\ne0$ , and either both $f,g\to0$ or both have infinite magnitude. If $f'/g'$ has a finite or infinite limit $L$ , then $f/g\to L$ . Apply separately on both sides for a two-sided limit. This is a sufficient theorem; failure of the derivative-ratio limit does not prove that the original limit fails. Check the indeterminate form again before each repetition. [O1LH]

Forms $0\cdot\infty$ , $\infty-\infty$ , $1^\infty$ , $0^0$ , and $\infty^0$ require rewriting, not direct differentiation of both visible pieces. For variable powers with a positive base, use $y=u(x)^{v(x)}$ and $\ln y=v\ln u$ .

D. Why Newton’s formula and the limit rule work#

The tangent at $x_n$ is $y=f(x_n)+f'(x_n)(x-x_n)$ . Setting $y=0$ gives the iteration. This uses the tangent’s root as a guess for the curve’s root, so a poor local model can produce a poor next guess.

For a simple $0/0$ case at a finite endpoint, the Cauchy Mean Value Theorem relates a quotient of increments to a quotient of derivatives at an intermediate point. As the endpoint is approached, that intermediate point also approaches it, transferring a derivative-ratio limit to the function quotient. The general versions require more careful hypotheses, which is why the rule is not mere symbolic cancellation of zeros.

E. A decision procedure#

For a limit, try direct substitution first and identify its form. Factor, rationalize, or use a known limit if that resolves the structure transparently. Use L’Hôpital only after recording $0/0$ or $\infty/\infty$ and derivative conditions. For a power, take a logarithm on a positive-base neighborhood, find its limit, then exponentiate. For Newton, show the function, derivative, starting value, and actual iterates; check the final residual $f(x_n)$ and whether the iterate stayed in the domain. A small step is not alone a rigorous error bound.

F. Worked teaching examples#

A root iteration. To solve $x^2=5$ , use $f=x^2-5$ and $x_0=2$ . Then $x_{n+1}=(x_n+5/x_n)/2$ , so $x_1=9/4=2.25$ and $x_2=161/72\approx2.236111$ . The squared residual is $x_2^2-5=1/5184$ , not zero. An MVT derivative lower bound can convert this residual to a certified input error if the root and iterate are known to lie in a suitable interval.

An indeterminate power. For $x>0$ , let $y=(1+2x)^{1/x}$ . Then $\ln y=\ln(1+2x)/x$ , a $0/0$ quotient as $x\to0^+$ . Its derivative ratio is $2/(1+2x)\to2$ , so $y\to e^2$ . The exponent was not simply replaced by infinity.

A rule that does not apply. The limit $(x+1)/x$ as $x\to0^+$ is $+\infty$ , not $0/0$ . Differentiating numerator and denominator to obtain $1$ would be an unjustified and wrong conclusion.

From the engine: NightjarNewton’s method finds a satellite

Nightjar, the author’s satellite-tracking lab, needs a satellite’s position at a given time. On an elliptical orbit the angle $E$ that locates it solves Kepler’s equation $M = E - e\sin E$ , where $M$ grows steadily with time and $e$ is the eccentricity. No formula gives $E$ , so the code applies Newton’s method to $f(E) = E - e\sin E - M$ , whose derivative is $f'(E) = 1 - e\cos E$ .

Nightjar: two-body propagationC++

fate::gnc::State keplerState(const KeplerElements& k, const fate::gnc::Instant& t, double mu) {
    const double a = k.semiMajorAxisM;
    const double e = k.eccentricity;
    const double n = std::sqrt(mu / (a * a * a));
    const double m = std::remainder(k.meanAnomalyRad + n * t.secondsSince(k.epoch), kTwoPi);
    // Kepler's equation by Newton from E = M + e sin M.
    double big = m + e * std::sin(m);
    for (int i = 0; i < 50; ++i) {
        const double step = (big - e * std::sin(big) - m) / (1.0 - e * std::cos(big));
        big -= step;
        if (std::fabs(step) < 1e-15) break;
    }
    const double cosE = std::cos(big);
    const double sinE = std::sin(big);
    const double root = std::sqrt(1.0 - e * e);
    const double r = a * (1.0 - e * cosE);
    const double x = a * (cosE - e);
    const double y = a * root * sinE;
    const double vx = -std::sqrt(mu * a) / r * sinE;
    const double vy = std::sqrt(mu * a) / r * root * cosE;
    const double cO = std::cos(k.raanRad);
    const double sO = std::sin(k.raanRad);
    const double cw = std::cos(k.argPerigeeRad);
    const double sw = std::sin(k.argPerigeeRad);
    const double ci = std::cos(k.inclinationRad);
    const double si = std::sin(k.inclinationRad);
    const Vec3 p{cO * cw - sO * sw * ci, sO * cw + cO * sw * ci, sw * si};
    const Vec3 q{-cO * sw - sO * cw * ci, -sO * sw + cO * cw * ci, cw * si};
    fate::gnc::State s;
    s.position = p * x + q * y;
    s.velocity = p * vx + q * vy;
    return s;
}

Take $e = 0.1$ and $M = 1$ radian. The starting guess $E_0 = M + e\sin M = 1.08415$ leaves $|f| = 4.2\times 10^{-3}$ . One step gives $E_1 = 1.08860$ and $|f| = 8.8\times 10^{-7}$ ; a second step leaves less than $10^{-11}$ . The number of correct digits roughly doubles with each step. That is Newton’s quadratic convergence, the rapid convergence this module describes near a simple root, and the root here is always simple: $f' \ge 1 - e > 0$ on every closed orbit. The lines after the loop turn $E$ into a position and a velocity in the orbit’s plane and rotate them into space.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M15.01FSolution

Take two Newton steps for $f(x)=x^2-3$ starting at $x_0=2$ . Keep exact fractions and evaluate the second iterate’s residual.

M15.02TSolution

Show that Newton’s method for $f(x)=x^3-2x+2$ starting at $0$ cycles. Does this prove the equation has no real root?

M15.03CSolution

Evaluate $\lim_{x\to0}(e^{3x}-1)/x$ . State why L’Hôpital applies.

M15.04CSolution

Evaluate $\lim_{x\to0}(\sin x-x)/x^3$ .

M15.05CSolution

Evaluate $\lim_{x\to\infty}x^2/e^x$ and interpret the growth comparison.

M15.06TSolution

Evaluate $\lim_{x\to0^+}x\ln x$ by converting it to a quotient.

M15.07TSolution

Evaluate $\lim_{x\to0^+}(1+3x)^{2/x}$ .

M15.08SSolution

A root $r$ and approximation $a$ both lie in an interval where $f$ is continuous, differentiable in its interior, and $|f'|\ge m>0$ . Prove $|a-r|\le|f(a)|/m$ . Apply it to $a=7/5$ for the root of $x^2-2$ in $[1,2]$ .

Optional small hints

For M15.04, check the form after each derivative. For M15.06, write $\ln x/(1/x)$ . For M15.08, apply the MVT between the root and the approximation, not automatically across the entire interval.

H. Complete solutions: Module 15#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M15.01Show solution

Exact iterates and a nonzero residual

Here $f'=2x$ , so $x_{n+1}=(x_n+3/x_n)/2$ . Therefore $x_1=(2+3/2)/2=7/4$ and $x_2=(7/4+12/7)/2=\boxed{97/56}$ . Its residual is $(97/56)^2-3=(9409-9408)/3136=\boxed{1/3136}$ . The small positive residual indicates that this positive approximation is slightly above $\sqrt3$ .

Back to M15.01

M15.02Show solution

A two-cycle is a failure of this starting guess

Since $f'=3x^2-2$ , $x_1=0-2/(-2)=1$ . At $1$ , $f(1)=1$ and $f'(1)=1$ , so $x_2=1-1=0$ . The same two steps repeat forever. Nevertheless $f(-2)=-2$ and $f(-1)=3$ ; polynomial continuity and the IVT guarantee a root between them. An algorithm’s failure does not establish nonexistence.

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M15.03Show solution

Check the form, then differentiate

Both numerator and denominator tend to zero; they are differentiable and the denominator derivative is $1\ne0$ . The derivative ratio $3e^{3x}$ tends to $3$ , so the requested limit is $\boxed3$ . The original quotient is not equal to $3e^{3x}$ for general nonzero $x$ .

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M15.04Show solution

Repeated use requires repeated checks

The first quotient is $0/0$ . L’Hôpital transforms its limit into that of $(\cos x-1)/(3x^2)$ , still $0/0$ . A second use gives $-\sin x/(6x)$ , again $0/0$ . A third gives $-\cos x/6\to\boxed{-1/6}$ . The denominator derivatives are nonzero on sufficiently small punctured intervals, and the final limit exists. A Taylor expansion later will independently explain the cubic leading term.

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M15.05Show solution

Polynomial growth versus exponential growth

The form is $\infty/\infty$ . Differentiating gives $2x/e^x$ , still $\infty/\infty$ ; repeating gives $2/e^x\to0$ . Thus $\boxed0$ : the quadratic is negligible relative to the exponential as $x\to\infty$ . This is an eventual comparison, not a claim that $e^x>x^2$ at every real input.

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M15.06Show solution

Turn a product into a valid quotient

For $x>0$ , $x\ln x=\ln x/(1/x)$ , of infinite-over-infinite magnitude as $x\to0^+$ . The derivative ratio is $(1/x)/(-1/x^2)=-x\to0$ . Hence the original limit is $\boxed0$ , approached through negative values. Substituting “zero times negative infinity” is not arithmetic.

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M15.07Show solution

Take logarithms without losing the final step

The base is positive near zero. Set $y=(1+3x)^{2/x}$ . Then $\ln y=2\ln(1+3x)/x\to2(3)=6$ by a valid $0/0$ application. Continuity of the exponential gives $\boxed{y\to e^6}$ . Reporting $6$ would give the logarithm’s limit, not the requested power’s limit.

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M15.08Show solution

A residual can become a rigorous error bound

If $a=r$ , the result is immediate. Otherwise the MVT gives $f(a)-f(r)=f'(c)(a-r)$ for some point between them. Since $f(r)=0$ , $|f(a)|\ge m|a-r|$ , proving the bound. For $f=x^2-2$ on $[1,2]$ , $f'=2x\ge2$ and $f(7/5)=-1/25$ . Consequently $\boxed{|7/5-\sqrt2|\le1/50=0.02}$ . This is a guaranteed bound from stated assumptions, unlike an unexplained decimal agreement.

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I. Retention and readiness#

Review M06.03, M03.06, and M13.06. Before every L’Hôpital calculation, say the indeterminate form aloud. Before trusting a numerical root, check the residual and the domain. Advance when you can reject a tempting but inapplicable rule as confidently as you can use it.