Question

2. If $(x) = e2x + 1, when x>0
1 + cos2x, when x < 0
show that, lim 0(x) = 2.
x-0​

Answer 1

Step-by-step explanation:

we have that

$f(x) = \begin{cases} e^{2x} +1,& \quad x >0,\\ 1+\cos 2x,& \quad x<0.\end{cases}$

We want to find the value of $\lim_{x\to 0} f(x)$.

As $f(x)$ is piece wise defined we need to check that

$\lim_{x\to0^-} f(x) = \lim_{x\to0^+} f(x)$.

First we compute

$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} e^{2x}+1 = e^{2(0)}+1 = 1+1 = 2$.

We then compute

$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} 1+ \cos 2x = 1+\cos 2(0) = 1+1 = 2$.

Then we have that

$\lim_{x\to0^-} f(x) = 2 = \lim_{x\to0^+} f(x)$,

which implies that

$\lim_{x\to 0} f(x) = 2$.