One sided limits Proof

tbyou87 · #1 $Old$ October 21st, 2007, 12:58 PM

Hi,
Let s,t ele R and f1,f2: (s,t) -> R. Suppose lim x->a- f1(x) = L1 and lim x->a- f2(x) = L2 with L1, L2 ele R.

Show that L = lim x->a- f1(x) + f2(x) exists and L = L1 + L2.

I know you need to use limits of sequences but I was not sure how to do it with one sided limits.

Thanks

Jhevon · #2 $Old$ October 21st, 2007, 01:14 PM

Quote:

Originally Posted by tbyou87 $View Post$

Hi,
Let s,t ele R and f1,f2: (s,t) -> R. Suppose lim x->a- f1(x) = L1 and lim x->a- f2(x) = L2 with L1, L2 ele R.

Show that L = lim x->a- f1(x) + f2(x) exists and L = L1 + L2.

I know you need to use limits of sequences but I was not sure how to do it with one sided limits.

Thanks

$\lim_{x \to a^-}f_1(x) = L_1$ means $f_1$ is defined on $S \subset \mathbb{R}$ , where $S = \{ x : x \in \mathbb{R}, x < a \}$ , and for any sequence $\{ x_n \}$ in $S$ with limit $a$ , (all $x_n < a$ ), we have:

$\lim_{n \to \infty}f_1(x_n) = L_1$

A similar thing holds true for $f_2(x)$

Thus, $\lim_{x \to a^-} (f_1(x) + f_2(x)) = \lim_{n \to \infty} (f_1(x_n) + f_2(x_n)) = \lim_{n \to \infty}f_1(x_n) + \lim_{n \to \infty}f_2(x_n) = L_1 + L_2 = L$