A limit via asymptotic equivalence¶

Math I: Calculus · Mock exam, Question 4(b) · 14 points · Lecture 8

Problem¶

Define

\begin{aligned} N(x)&=(1+x)^{1/x}+\mathrm{e}^x-\mathrm{e}-1,\\ D(x)&=\ln\left(x+\sqrt{1+x^2}\right). \end{aligned}

(b1) Prove that $D(x)\sim x$ as $x\to0\text{.}$ [4 points]
(b2) Evaluate $\lim_{x\to0}N(x)/D(x)\text{.}$ [10 points]

What you need¶

Asymptotic equivalence: $f(x)\sim g(x)$ means $f(x)/g(x)\to1\text{.}$
Derivative and differentiability: if $F$ is differentiable at $0$ and $F(0)=0\text{,}$ then $F(x)/x\to F'(0)\text{.}$
Chain rule, logarithm and exponential function.
Indeterminate form and L'Hôpital's rule.
Euler's number: $(1+x)^{1/x}\to\mathrm{e}$ as $x\to0\text{.}$

Plan¶

Split the hard limit into pieces that are each a known difference quotient. Part (b1) lets you replace $D(x)$ by $x\text{.}$ Then $N(x)/x$ splits into $[(1+x)^{1/x}-\mathrm{e}]/x$ and $(\mathrm{e}^x-1)/x\text{.}$ The first piece is the only real work: write $(1+x)^{1/x}=\mathrm{e}^{u(x)}$ and use the chain of quotients $\frac{\mathrm{e}^{u}-\mathrm{e}}{u-1}\cdot\frac{u-1}{x}\text{.}$

Step-by-step solution¶

Try each step yourself before you open it.

(b1) Step 1 · Connect the limit with D′(0) (1 point)

$D(0)=\ln1=0\text{,}$ so

\frac{D(x)}{x}=\frac{D(x)-D(0)}{x-0}\to D'(0).

It is enough to show $D'(0)=1\text{.}$

(b1) Step 2 · Differentiate and simplify (2 points)

By the chain rule,

\begin{aligned} D'(x)&=\frac{1+x/\sqrt{1+x^2}}{x+\sqrt{1+x^2}}\\ &=\frac{(\sqrt{1+x^2}+x)/\sqrt{1+x^2}}{x+\sqrt{1+x^2}}\\ &=\frac{1}{\sqrt{1+x^2}}. \end{aligned}

(b1) Step 3 · Conclude (1 point)

$D'(0)=1\text{,}$ so $D(x)/x\to1\text{,}$ that is, $D(x)\sim x$ as $x\to0\text{.}$

Concept: Asymptotic equivalence

(b2) Step 1 · Rewrite the power (2 points)

Let $u(x)=\ln(1+x)/x$ for $x\neq0\text{.}$ Then $(1+x)^{1/x}=\mathrm{e}^{u(x)}\text{,}$ and $u(x)\to1$ because $\ln(1+x)/x$ is the difference quotient of $\ln(1+x)$ at $0\text{.}$

(b2) Step 2 · The rate at which u approaches 1 (3 points)

\begin{aligned} \lim_{x\to0}\frac{u(x)-1}{x} &=\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}\\ &=\lim_{x\to0}\frac{1/(1+x)-1}{2x}\\ &=\lim_{x\to0}\frac{-1}{2(1+x)}=-\frac{1}{2}. \end{aligned}

The second equality is L'Hôpital's rule for the form $0/0\text{;}$ it is justified because the last limit exists.

Concept: L'Hôpital's rule

(b2) Step 3 · The first piece of N(x)/x (2 points)

For small $x\neq0\text{,}$ $u(x)\neq1$ because $\ln(1+x)<x\text{.}$ So we may write

\frac{(1+x)^{1/x}-\mathrm{e}}{x} =\frac{\mathrm{e}^{u(x)}-\mathrm{e}}{u(x)-1}\cdot\frac{u(x)-1}{x} \to\mathrm{e}\cdot\left(-\frac{1}{2}\right)=-\frac{\mathrm{e}}{2}.

The first factor tends to the derivative of $\mathrm{e}^t$ at $t=1\text{,}$ which is $\mathrm{e}\text{.}$

(b2) Step 4 · The second piece (1 point)

$N(x)=\left[(1+x)^{1/x}-\mathrm{e}\right]+\left[\mathrm{e}^x-1\right]\text{,}$ and $(\mathrm{e}^x-1)/x\to1\text{.}$ Hence

\frac{N(x)}{x}\to1-\frac{\mathrm{e}}{2}.

(b2) Step 5 · Use (b1) and conclude (2 points)

By (b1), $x/D(x)\to1\text{.}$ Therefore

\frac{N(x)}{D(x)}=\frac{N(x)}{x}\cdot\frac{x}{D(x)}\to1-\frac{\mathrm{e}}{2}.

Where marks are lost¶

Applying L'Hôpital's rule to $N(x)/D(x)$ directly. It is allowed in principle, but the derivative of $(1+x)^{1/x}$ is messy and contains another quotient of type $0/0\text{.}$ Splitting into known quotients is shorter.
Writing $(1+x)^{1/x}\approx\mathrm{e}$ and cancelling. That throws away exactly the term of order $x$ that the limit depends on.
Not citing (b1). The last step uses $x/D(x)\to1\text{;}$ say so.

Try next¶

The inverse of $x+x^3$ : another $0/0$ limit, solved in three ways.
Area between two curves and $\int_0^1\ln x\,\mathrm{d}x$

Symbols: integral, square root