## University Calculus: Early Transcendentals (3rd Edition)

Prove that for $f(x)=k$, then $\lim_{x\to\infty}f(x)=k$, which means to prove for every $\epsilon\gt0$, there exists a corresponding $M\gt0$ such that for all $x$ $$x\lt M\Rightarrow |f(x)-k|\lt\epsilon$$
*The formal definition of limits as $x\to\infty$: $\lim_{x\to\infty}f(x)=L$ is for every number $\epsilon\gt0$, there exists a corresponding number $M\gt0$ such that for all $x$ $$x\lt M\Rightarrow |f(x)-L|\lt\epsilon$$ Here we need to prove for $f(x)=k$, then $\lim_{x\to\infty}f(x)=k$, which means to prove for every $\epsilon\gt0$, there exists a corresponding $M\gt0$ such that for all $x$ $$x\lt M\Rightarrow |f(x)-k|\lt\epsilon$$ Examine the inequality: $$|f(x)-k|\lt\epsilon$$ Since $f(x)=k$: $$|k-k|\lt\epsilon$$ $$|0|\lt\epsilon$$ $$0\lt\epsilon$$ which is always true for all $x$. In other words, $|f(x)-k|\lt\epsilon$ is always true no matter which $M$ we choose to limit the values of $x$. Therefore, $\lim_{x\to\infty}f(x)=k$ as $f(x)=k$