Question

1. a). Write down the definition (ε-δ language) of limx→x0 f(x) = L. b). Show that limx→+∞ cos x does not exist.

Answer 1

a). Let us write down the definition ($ε-δ$ language) of $\lim\limits_{x→x_0} f(x) = L$:

$\lim\limits_{x→x_0} f(x) = L$ if and only if for any $\varepsilon>0$ there exists $\delta>0$ such that for any $x\ne x_0$ if $|x-x_0|<\delta$ then $|f(x)-L|<\varepsilon.$

b). Let us show that $\lim\limits_{x→+\infty} \cos x$  does not exist. Suppose contrary that such limit exists and equals to $L$. Then for any $\varepsilon>0$ there exists $\Delta>0$ such that for any $x>\Delta$ we have that $|f(x)-L|<\varepsilon.$

Let $\varepsilon =1$. Then for any even integer number $n>\Delta$ we have that $n\pi>\Delta$ and $n\pi+\pi>\Delta$. Since the period of the function $y=\cos x$ is $2\pi$, we conclude that $\cos n\pi =\cos 0 =1$ and $\cos (n\pi+\pi) =\cos \pi =-1$. By definition of limit we have that $|1-L|=|\cos n\pi -L|<1$ and $|-1-L|=|\cos (n\pi+\pi) -L|<1$. The latter inequalities are equivalent to $0<L<2$ and $-2<L<0,$ that is impossible. This contradiction proves that the limit does not exist.