- #1
Tsunoyukami
- 215
- 11
Suppose ##\lim_{x \to a} f(x) = \infty## and ##\lim_{x \to a} g(x) = c## where ##c## is a real number. Prove ##\lim_{x \to a} \big( f(x)+g(x) \big) = \infty##.
Proof Assume ##\lim_{x \to a} f(x) = \infty## and ##\lim_{x \to a} g(x) = c##. Then, by definition:
1) For every ##M > 0## there exists ##\delta_{1} > 0## such that ##0<|x-a|<\delta_1 \implies f(x)>M##.
2) For every ##\epsilon >0## there exists ##\delta_{2} >0## such that ##0<|x-a|<\delta_{2} \implies |g(x)-c|<\epsilon##.
We want to show that for every ##S>0## there exists ##\delta>0## such that ##0<|x-a|<\delta \implies f(x)+g(x)>S##.
Choose ##\delta = \min(\delta_{1}, \delta_{2})##.
Then ##0<|x-a|<\delta \implies f(x)>M## and Then ##0<|x-a|<\delta \implies |g(x)+c|<\epsilon \implies c -\epsilon < g(x) < c + \epsilon##.
Then ##f(x) + g(x) > M - \epsilon + c = S## and so, with ##S = M + c - \epsilon## we conclude ##\lim_{x \to a} \big( f(x)+g(x) \big) = \infty##.
Is this a valid proof? I'm not entirely sure if my argument is valid near the end. Any help is appreciated! Thanks.
Proof Assume ##\lim_{x \to a} f(x) = \infty## and ##\lim_{x \to a} g(x) = c##. Then, by definition:
1) For every ##M > 0## there exists ##\delta_{1} > 0## such that ##0<|x-a|<\delta_1 \implies f(x)>M##.
2) For every ##\epsilon >0## there exists ##\delta_{2} >0## such that ##0<|x-a|<\delta_{2} \implies |g(x)-c|<\epsilon##.
We want to show that for every ##S>0## there exists ##\delta>0## such that ##0<|x-a|<\delta \implies f(x)+g(x)>S##.
Choose ##\delta = \min(\delta_{1}, \delta_{2})##.
Then ##0<|x-a|<\delta \implies f(x)>M## and Then ##0<|x-a|<\delta \implies |g(x)+c|<\epsilon \implies c -\epsilon < g(x) < c + \epsilon##.
Then ##f(x) + g(x) > M - \epsilon + c = S## and so, with ##S = M + c - \epsilon## we conclude ##\lim_{x \to a} \big( f(x)+g(x) \big) = \infty##.
Is this a valid proof? I'm not entirely sure if my argument is valid near the end. Any help is appreciated! Thanks.