If the limit of f(x) as x tends to a = L and the limit of g(x) as x tends to a =M Proof that, lim[f(x)-g(x)] as x tends to a = L-M?

As it is written the statement is wrong.

Jan 24, 2018

We have to use the limit definition

Explanation:

Because of the limit definition we know:

${\lim}_{x \to c} h \left(x\right) = K$ iff given any $\epsilon > 0$ there exists $\delta$ such that if $0 < | x - c | < \delta$ $\implies | h \left(x\right) - K | < \epsilon$

Now let's consider any $\epsilon > 0$:

Since ${\lim}_{x \to a} f \left(x\right) = L$, given $\frac{\epsilon}{2} > 0$ there exists ${\delta}_{1}$ such that if $0 < | x - a | < {\delta}_{1}$ $\implies | f \left(x\right) - L | < \frac{\epsilon}{2}$

Since ${\lim}_{x \to a} g \left(x\right) = M$, given $\frac{\epsilon}{2} > 0$ there exists ${\delta}_{2}$ such that if $0 < | x - a | < {\delta}_{2}$ $\implies | g \left(x\right) - M | < \frac{\epsilon}{2}$

If we now take $\delta = \min \left({\delta}_{1} , {\delta}_{2}\right)$, we can say:

$0 < | x - a | < \delta$ $\implies$ $0 < | x - a | < {\delta}_{1}$ and $0 < | x - a | < {\delta}_{2}$

and therefore:

if $0 < | x - a | < \delta \implies$

$| \left(f \left(x\right) - g \left(x\right)\right) - \left(L - M\right) | < | \left(f \left(x\right) - L\right) - \left(g \left(x\right) - M\right) | \le | f \left(x\right) - L | + | g \left(x\right) - M | < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

That is, the ${\lim}_{x \to a} \left[f \left(x\right) - g \left(x\right)\right] = L - M$