Limits

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

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.

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

(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 \(f(x)=2016\)

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 \(x \geq 0 \).

(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 \(t=1 / h\)