Prove the existence of two limits I am trying to solve two basic problems of limits of functions in one and two variables, the exercises are the following:
1) Let $f:[0,+\infty) \to \mathbb R$ be a continuous and positive function such that $\lim_{x \to +\infty} f(x)=a>0$. Prove that $\lim_{x \to +\infty} \frac{1}{x} 
\int_0^x f(t)dt=a$
2) Analyze the existence of the limit at the point $(0,0)$ of the function $f(x,y)=x\ln(x^2+y^2)$
My attempt at a solution:
First of all, I realize I must grab Spivak, Apostol, Courant & company textbooks and read and work through the exercises since I can't prove basic single-variable statements but, until that, what I've done is:
For 1) could it be the case that I have to use the fundamental theorem of calculus?. I know that for a closed interval $[0,x_0]$, if I call $F(x)=\int_0^x f(t)dt$, then $F'(x)=f(x)$. Then $lim_{x \to +\infty} F'(x)=lim_{x \to +\infty} f(x)=a$. So, I suppose I have to show that $F'(x)=\frac{1}{x} \int_0^x f(t)dt$. Am I correct?
For 2), I think that the limit exists and is $0$. I am trying to formally prove this, i.e., let $\epsilon>0$, I want to show there is $\delta>0$: $\|(x,y)-(0,0)\|<\delta \implies |f(x,y)-0|<\epsilon$.
So, $|f(x,y)-0|=|x\ln(x^2+y^2)|=|x||\ln(x^2+y^2)|\leq \|(x,y)\||\ln(x^2+y^2)|$. How can I find an upper bound for $|\ln(x^2+y^2)|$?
 A: PART1: Let $f(x)$ be continuous on $[0,\infty)$ and approach a limit of $a$ as $x\rightarrow\infty$. Notice that
$$
    \frac{1}{x}\int_{0}^{x}f(t)\,dt - a = \frac{1}{x}\int_{0}^{x}(f(t)-a)\,dt.
$$
For any fixed $L$, and $x > L$,
$$
\frac{1}{x}\int_{0}^{x}f(t)\,dt - a = \frac{1}{x}\int_{0}^{L}(f(t)-a)\,dt +\frac{1}{x}\int_{L}^{x}(f(t)-a)\,dt.
$$
Let $\epsilon > 0$ be given. Because $\lim_{x\rightarrow\infty}f(x)=a$, then there exists $R > 0$ such that $|f(x)-a| < \epsilon/2$ whenever $x > R$. Let $L=R$. Then the second integral on the right is bounded by $\frac{x-L}{x}\frac{\epsilon}{2} < \frac{\epsilon}{2}$ whenever $x > L$. For this fixed $L$, the first integral term on the right is a constant times $1/x$ and, hence, tends to 0 as $x\rightarrow\infty$. So there exists $S > 0$ such that the first term on the right is bounded by $\epsilon > 0$ whenever $x > S$. Therefore, if $x > \max\{L,S\}$, one sees that the right side is strictly bounded by $\epsilon/2+\epsilon/2=\epsilon$. Because $\epsilon > 0$ was arbitrary, it follows that the limit of the above is 0 as $x\rightarrow\infty$.
PART2: For this part, you want a bound on $u\ln u$ for $0 < u < 1$. Just take the limit
$$
   \lim_{u\downarrow 0}u\ln u = \lim_{u\downarrow 0}\frac{\ln u}{(1/u)}=\lim_{u\downarrow 0}\frac{(1/u)}{(-1/u^{2})}=-\lim_{u\downarrow}u = 0.
$$
So, as you pointed out $|x\ln(x^{2}+y^{2})| \le \|(x,y)\|\ln \|(x,y)\|\rightarrow 0$ as $\|(x,y)\|\rightarrow 0$.
