Section Assignments # 3 & #4- Due Tuesday 2/11
- (Some of you finished this in class. If you did, just move on to the next problem.)
Consider the function \(f:[-1,1]\rightarrow \mathbb{R}\) by
\[
f(x)=
\begin{cases}
1+2x^2, & x\; \textrm{rational}\\
1+x^4, & x\; \textrm{irrational}.
\end{cases}
\]
(Recall a rational number is a number that can be written as a quotient of integers.)
Find \(\lim_{x\rightarrow 0} f(x)\) and clearly justify your answer.
The difficult thing about this problem is that as \(x\rightarrow 0\) it passes through both rational and irrational numbers. That is, no matter how small an open interval we consider around \(x=0\), it will always contain both rational and irrational numbers. So, the rule used to assign the value \(f(x)\) is different for different numbers for every interval around 0. The key is to sort of get rid of this difficulty. One way to do this is simply bound \(f(x)\) and see what this tells us about its limit.
Since both \(2x^2\) and \(x^4\) are non-negative, we know \( 1 \leq f(x)\) for all \(x\). Also, in the interval [-1,1], \(x^4 \leq x^2 \leq 2 x^2 \). From this we conclude \(f(x) \leq 1+2x ^2\) for all \(x \in [-1,1]\). This implies
\[
\lim_{x\rightarrow 0} 1 \leq \lim_{x\rightarrow 0} f(x) \leq \lim_{x\rightarrow 0} 1+2x^2,\;\; \textrm{for all}\; x \in [-1,1].
\]
Since \(\lim_{x\rightarrow 0} 1=\lim_{x\rightarrow 0}1+2x^2=1\), by the Sandwich Theorem, we conclude \(\lim_{x\rightarrow 0} f(x)=1\).
-
Consider the functions \(f(x)=1\) and \(g(x)=\begin{cases}2x, & x \neq 1\\ 5, & x=1. \end{cases}\)
- Calculate \(\lim_{x \rightarrow 0} f(x)\).
\[
\lim_{x\rightarrow 0} f(x)=\lim_{x\rightarrow 0} 1=1.
\]
- Calculate \(\lim_{x \rightarrow 1} g(x)\).
\[
\lim_{x\rightarrow 1}g(x) =\lim_{x\rightarrow 1} 2x =2.
\]
- Find an expression for the composition \(g\circ f (x)=g(f(x))\).
\[
g(f(x))=g(1)=5,\;\textrm{for all}\; x.
\]
- Using your answer to part c, calculate \(\lim_{x \rightarrow 0} g(f(x))\).
\[
\lim_{x\rightarrow 0} g(f(x))=\lim_{x\rightarrow 0} 5=5.
\]
- Based on your above answers, do you think if we have two functions \(h(x)\) and \(j(x)\) which statisfy \(\lim_{x\rightarrow c}h(x)=L\;\) and \(\;\lim_{x\rightarrow L}j(x)=M,\;\) then it is safe to say that \(\;\lim_{x\rightarrow c}j(h(x))=M\,\)?
In our example, \(\lim_{x\rightarrow 0} f(x)=1\), and \(\lim_{x\rightarrow 1} g(x)=2\). However, from part d we know \(\lim_{x\rightarrow 0}g(f(x))=5,\) not 2. So no, we shouldn't think this is true in general.