I'm trying to prove that if $\lim\limits_{x \to a} f_i(x) = L_i$ for each $i=1,2,\ldots, n$, then $$\lim_{x\to a} \big(f_1(x)+f_2(x)+ \cdots +f_n(x)\big) = L_1+L_2+ \cdots+L_n$$ I've been asked to prove this result by induction, which I am awfully rusty at.
Could anyone give me a hint/motivation for how I should approach this.
My current attempt has been substituting $n$ in the last line for $(k+1)$ (assuming the line is true for $n=k$) but I just lost myself in the algebra of this. Any help would be very much appreciated. Thanks.
Maybe you should try and brush up on your induction first- but here is a sketch of the proof.
Base Case: clearly the statement holds if $n=1$, since this is equivalent to $\lim_{x\to a}f_1(x)=L_1$, which we are given.
Inductive step Assume the statement holds for some $n=k\in \mathbb{N}$, that is, if $\{f_1,...,f_k\}$ is any collection of $k$ functions such that $\lim_{x\to a}f_i(x)=L_i$ for $1\leq i\leq k$, then: $$\lim_{x\to a}[f_1(x)+...+f_k(x)]=L_1+...+L_k. $$
Now suppose we have $k+1$ functions $\{f_1,...,f_{k+1}\}$ such that $\lim_{x\to a}f_i(x)=L_i$ for $1\leq i\leq k+1.$ Then:
\begin{align} \lim_{x\to a}[f_1(x)+...+f_k(x)+f_{k+1}]&=\lim_{x\to a}[f_1(x)+...+f_k(x)]+\lim_{x\to a}f_{k+1}(x) \quad (I)\\ &=(L_1+...+L_k)+L_{k+1} \quad (II)\\ &=L_1+...+L_k+L_{k+1} \end{align}
As desired. Thus by induction we are done.
In the first line $(I)$ I have used that if $f,g$ are two functions such that $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist, then $\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$, where in this specific case I have used $f:=f_1+...+f_k,\enspace g:=f_{k+1}$. In order to establish $(II)$, I have used that We know the limit of $f(=f_1+...+f_k)$ as $x\to a$ exists and equals $L_1+...+L_k$ by the induction assumption.