# Limits

### Vanilla application of L’Hospital

A3, 2010

Evaluate the limit:

$\lim_{x \rightarrow 1}\left(\frac{n- \displaystyle \sum_{k=1}^n x^{k}}{1-x}\right)$

Solution We apply the L'Hospital's rule and differentiate both the numerator and the denominator. \begin{array}{rl} \lim_{x\rightarrow 1}&\displaystyle \frac{-n x^{n-1}-(n-1) x^{n-2}-\cdots-x-1}{-1} \\ =&\frac{n(n-1)}{2} \end{array}

Practice Problem

Explain what is wrong with the use of L'Hospital's rule. Find the correct limit. $\lim_{x \rightarrow 1} \frac{2 x^{3}-3 x+1}{x^{4}-1}=\lim_{x \rightarrow 1} \frac{6 x^{2}-3}{4 x^{3}}=\lim_{x \rightarrow 1} \frac{12 x}{12 x^{2}}=\lim_{x \rightarrow 1} \frac{1}{x}=1$

### L’Hospital again

A4, 2012

Prove the following limit.

$\lim_{x \rightarrow \infty} \frac{x^{100} \ln (x)}{e^{x} \arctan \left(\frac{\pi}{3}+\sin x\right)}=0$

Solution

For some positive constant $$k$$ we can ensure that $$\arctan \left(\frac{\pi}{3}+\sin x\right)> k$$ for any $$x$$. For example, $$k=\arctan(0.0001)$$ will work. This is because $$\pi>3.142, \sin (x) \geq-1$$ and $$\arctan$$ is an increasing function.

Further, $$\ln (x)< x$$ for $$x> 0$$. So the given ratio must lie between 0 and $$x^{101} / c e^{x}$$. If we apply the L'Hospital's rule 102 times, we get the result.

### Limit of $$x^{x^{x}}$$

B1a, 2017

(a) Evaluate $$\lim_{x \rightarrow 0^{+}}\left(x^{x^{x}}-x^{x}\right)$$

Solution

First consider the limit \begin{align} \lim_{x \rightarrow 0^{+}} x^{x} \\ &=\lim_{x \rightarrow 0^{+}}\left(e^{x\log x}\right) \\ &=\lim_{x \rightarrow 0^{+}}\left(e^{\frac{\log x}{1 / x}}\right) \end{align}

Now consider just the exponent: \begin{align} \lim_{x \rightarrow 0^{+}} \frac{\log x}{1 / x} \\ &=\lim_{x \rightarrow 0} \frac{1 / x}{-1 / x^{2}} \\ &=0 \end{align} substituting the value 0 from (2) in equation (1) we get that the limit is 1.

Now \begin{align} \lim_{x \rightarrow 0^{+}}\left(x^{x^{x}}-x^{x}\right) \\ &=\lim_{x \rightarrow 0^{+}} x^{x^{x}}-\lim_{x \rightarrow 0^{+}} x^{x} \\ &=\lim_{x \rightarrow 0^{+}} x^{\lim_{x \rightarrow 0^{+}} x^{x}}-\lim_{x \rightarrow 0^{+}} x^{x} \\ &=0-1 \\ &=-1 \end{align}

Reference. This problem is based on a general result: A tower of even number of $$x$$s tends to zero and a tower of odd number of $$x$$s tends to one .
 The limit of x** as x tends to zero J. Marshall Ash, Mathematics Magazine, 69 (1996), 207-209.

### Simple limits

A9, 2021

Let $$f$$ and $$g$$ be function defined as follows:
$f(x) = \frac{x}{x+\sin x}$ $g(x) = \frac{x^4+x^6}{e^x-1-x^2}$ (a) $$\lim_{x\rightarrow 0} f(x) = 1/2$$
(b) $$\lim_{x\rightarrow \infty} f(x)$$ does not exist.
(c) $$\lim_{x\rightarrow \infty} g(x)$$ is finite.
(d) $$\lim_{x\rightarrow 0} g(x) = 720$$.

Solution (a) True. $$\sin x \approx x$$ as $$x\rightarrow 0$$.
(b) False. The limit is 1.
(c) True. Numerator is polynomial, while the denominator is exponential.
(d) False. The limit is zero. (L'Hosptial's rule is not applicable).

### Smallest prime factor function

A9, 2017

Let $$f$$ be a continuous function from $$\mathbb{R}$$ to $$\mathbb{R}$$ (where $$\mathbb{R}$$ is the set of all real numbers) that satisfies the following property:

For every natural number $$n$$

$f(n)= \text{the smallest prime factor of }n$

For example, $$f(12)=2, f(105)=3 .$$ Calculate the following.

(a) $$\lim_{x \rightarrow \infty} f(x)$$

(b) The number of solutions to the equation $$f(x)=2016$$

Solution

(a) $$f(x)$$ will take value 2 for all even $$x$$. At the same time, primes provide an increasing infinite sequence of positive integers for which $$f(x)=x .$$ Thus $$\lim_{x \rightarrow \infty} f(x)$$ does not exist.

(b) By intermediate value theorem, for each prime $$p> 2016$$ there is an $$x$$ between $$p$$ and $$p+1$$ such that $$f(x)=2016$$

### Limit of a log of an exponent

A9, 2019

Consider $$f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$$ defined as follows:

$f(a, b):=\lim_{n \rightarrow \infty} \frac{1}{n} \log_{e}\left[e^{n a}+e^{n b}\right]$

(a) $$f$$ is not onto i.e. the range of $$f$$ is not all of $$\mathbb{R}$$.

(b) For every $$a$$ the function $$x \mapsto f(a, x)$$ is continuous everywhere.

(c) For every $$b$$ the function $$x \mapsto f(x, b)$$ is differentiable everywhere.

(d) We have $$f(0, x)=x$$ for all $$x \geq 0$$

Solution

Without loss of generality, assume that $$a > b$$.

\begin{align*} f(a, b) &=\lim_{n \rightarrow \infty} \frac{1}{n} \log\left(e^{n a}+e^{n b} \right) \\ & =\lim_{n \rightarrow \infty} \frac{1}{n} \log \left( e^{na}( 1 +e^{n(b-a)} ) \right) \\ & =\lim_{n \rightarrow \infty} \frac{1}{n} \log e^{na} + \log ( 1 +e^{n(b-a)} )\\ & = a + \lim_{n\rightarrow \infty} \frac{1}{n} \log ( 1 + e^{-\infty} ) \\ & = a + \lim_{n\rightarrow \infty} \frac{1}{n} \log ( 1 )\\ & = a \end{align*}

Similary, if $$b \geq a$$, $$f (a,b) = b$$. Thus, we have: $f(a,b) = \max(a,b)$

1. False. $$f(x,x) = x$$ so $$f$$ is onto.
2. True. $$g(x) = \max(x,a)$$ which is continuous everywhere.
3. False. $$h(x) = \max(x,b)$$ is not differentiable at $$x = b$$. More explicitly, $$\lim_{x \rightarrow b^+} = 1$$ and $$\lim_{x \rightarrow b^-} = 0$$.
4. True. $$\max(x,0) = x$$ for all $$x \geq 0$$.

### Polynomial and limits

B1, 2015

(a) For any polynomial $$p(t),$$ the limit $$\lim_{t \rightarrow \infty} \frac{p(t)}{e^{t}}$$ is independent of the polynomial $$p .$$ Justify this statement and find the value of the limit.

(b) Consider the function defined by

\begin{align} q(x) &=e^{-1 / x} \text { when } x>0 \\ &=0 \text { when } x=0 \\ &=e^{1 / x} \text { when } x<0 \end{align}

Show that $$q^{\prime}(0)$$ exists and find its value. Why is it enough to calculate the relevant limit from only one side?

(c) Now for any positive integer $$n,$$ show that $$q^{(n)}(0)$$ exists and find its value. Here $$q(x)$$ is the function in part (b) and $$q^{(n)}(0)$$ denotes its $$n$$ -th derivative at $$x=0$$.

Solution

(a) If $$p(t)$$ is constant, then the limit $$=0 .$$ Otherwise we get a form $$\frac{\pm \infty}{\infty}$$. Using L'Hospital's rule, we get $$\lim_{t \rightarrow \infty} \frac{p(t)}{e^{t}}=\lim_{t \rightarrow \infty} \frac{p^{\prime}(t)}{e^{t}}=0$$ by induction on the degree of $$t$$ (or apply L'Hospital's rule repeatedly).

(b) The right side derivative $$=\lim_{h \rightarrow 0^{+}} \frac{q(h)-q(0)}{h}=\lim_{h \rightarrow 0^{+}} \frac{e^{-1 / h}}{h}=\lim_{h \rightarrow 0^{+}} \frac{1 / h}{e^{1 / h}}=\lim_{t \rightarrow+\infty} \frac{t}{e^{t}} \cdot$$ (Let $$t=1 / h .$$ As $$\left.h \rightarrow 0^{+}, t \rightarrow+\infty .\right)$$ This limit is $$0,$$ e.g. by part $$(\mathrm{a})$$ Now $$q$$ is an even function, so letting $$k=-h,$$ the left side derivative $$=\lim_{h \rightarrow 0^{-}} \frac{q(h)-q(0)}{h}=$$ $$\lim_{k \rightarrow 0^{+}} \frac{q(-k)}{-k}=\lim_{k \rightarrow 0^{+}} \frac{q(k)}{-k} .$$ Using the earlier calculation this also equals $$-0=0$$ Note: It is wrong to argue that $$q^{\prime}(0)=\lim_{x \rightarrow 0} q^{\prime}(x)$$ because to do so, we first need to know that $$q^{\prime}$$ is continuous at $$0,$$ but we have not even shown that $$q^{\prime}(0)$$ exists! For the same reason it is wrong to argue below that $$q^{(n)}(0)=\lim_{x \rightarrow 0} q^{(n)}(x)$$

(c) We will show by induction on $$n$$ that $$q^{(n)}(0)=0 .$$ The case $$n=1$$ is done. (We can even start the induction at $$n=0$$ by interpreting $$q^{(0)}(x)=q(x) .$$ ) Assuming that we are done up to $$n$$ and to prove the statement for $$n+1,$$ we need to calculate $$\lim_{h \rightarrow 0} \frac{q^{(n)}(h)-q^{(n)}(0)}{h}=$$ $$\lim_{h \rightarrow 0} \frac{q^{(n)}(h)}{h},$$ because $$q^{(n)}(0)=0$$ by induction. Therefore it is good to examine $$q^{(n)}(h)$$ for $$h \neq 0 .$$ This is easy to calculate by the usual rules, but the formulas will be different for positive and negative $$h .$$ For $$h \neq 0,$$ as $$q$$ is even, $$q^{\prime}$$ is odd, so $$q^{\prime \prime}$$ is even, etc. and in general $$q^{(n)}(h)=(-1)^{n} q^{(n)}(-h) .$$ Therefore, just as for part (b), it suffices to show that $$\lim_{h \rightarrow 0^{+}} \frac{q^{(n)}(h)}{h}=0 .$$ By another induction, we see easily that for $$h>0, q^{(n)}(h)=p(1 / h) e^{-1 / h}$$ for some polynomial $$p .\left[\right.$$ Proof: $$q^{\prime}(h)=\left(\frac{1}{h^{2}}\right) e^{-1 / h} .$$ Assuming the result for $$n,$$ we have $$q^{(n+1)}(h)=\left[p(1 / h) e^{-1 / h}\right]^{\prime}=-\frac{1}{h^{2}}\left(-p(1 / h)+p^{\prime}(1 / h)\right) e^{-1 / h},$$ which has the desired form.

So we have $$\lim_{h \rightarrow 0^{+}} \frac{q^{(n)}(h)}{h}=\lim_{h \rightarrow 0^{+}} \frac{p(1 / h) e^{-1 / h}}{h}=\lim_{t \rightarrow \infty} t p(t) e^{-t}=\lim_{t \rightarrow \infty} \frac{t p(t)}{e^{t}}=0$$ by part (a). Here we have again substituted $$t=1 / h$$