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

Using the definiton of limits to define $\lim_{h\to0}f(h+c)=L$, then substitute $h+c=x$ to deduce the definition for $\lim_{x\to c}f(x)=L$.
$\lim_{x\to c}f(x)=L$ if and only if $\lim_{h\to0}f(h+c)=L$ According to definition, $\lim_{h\to 0}f(h+c)=L$ means that for a value of $\epsilon\gt0$, there is always a corresponding value of $\delta$ such that for all $h+c$, $$0\lt|h|\lt\delta\Rightarrow |f(h+c)-L|\lt\epsilon$$ If we substitute $h+c=x$, then $h=x-c$, meaning that for a value of $\epsilon\gt0$, there is always a corresponding value of $\delta$ such that for all $x$, $$0\lt|x-c|\lt\delta\Rightarrow |f(x)-L|\lt\epsilon$$ which is the definition of $\lim_{x\to c}f(x)=L$ The proof has been completed.