I am taking a basic calculus class and my professor never explained this but I have this question on the homework:
Suppose $\lim \limits_{x \to b} f(x) = 7$ and $\lim \limits_{x\to b} g(x) = −3$
Find $\lim \limits_{x\to b}(f(x)+g(x))$
How do you solve a problem like this?
$\endgroup$ 23 Answers
$\begingroup$Use the following result:
If $\lim \limits_{x \to b} f(x) $ and $\lim \limits_{x \to b} g(x) $ exist, then
$$\lim_{x \to b} (f(x) + g(x)) = \lim_{x \to b}f(x) + \lim_{x \to b} g(x) $$
Sum of limit is the limit of the sum.
You might want to read up more on what operations can you perform with limits.
$\endgroup$ 2 $\begingroup$If you know that the function, f, is "continuous" then, from the definition of "continuous", $\lim_{x\to b} f(g(x))= f(\lim_{x\to b}g(x))$.
But the example you give is not a "function inside a function"- you have a linear combination of two functions. One of the very first "laws of limits" you should have learned is ""$\lim_{x\to b} (f(x)+ g(x))= \lim_{x\to b} f(x)+ \lim_{x\to b} g(x)$.
$\endgroup$ $\begingroup$Let $\epsilon>0$, there exists $\delta_f,\delta_g>0$ such that
$|x-b|\leq \delta_f$ implies $|f(x)-7|<\epsilon/2$ and
$|x-b|\leq \delta_g$ implies $|g(x)+3|<\epsilon/2$.
It follows that for all $x$ such that $|x-b|\leq \min\{\delta_f,\delta_g\}$, we have
$$|f(x)+g(x)-4|\leq |f(x)-7|+|g(x)+3|\leq \epsilon$$
which implies $\lim_{x\to b}(f(x)+g(x))=4$
