Question

A function f(x) is defined as $f(x) = \left\{ \begin{array}{ll} \frac{1}{1+e^{1/x}} & \quad if \ \ x \ne 0 \\ 0 & \quad if \ \ x =0 \end{array} \right$ is thefunction continuous at x = 0?

Answer 1

oh I just remembered oh i forgot

Answer 2

Answer:

The function f(x) is not continuous at $x=0$

Step-by-step explanation:

In fact, since $\lim_{x \to 0^+}e^{\frac{1}{x}} = \infty$, we have $\lim_{x \to 0^+} \frac{1}{1+e^{\frac{1}{x}}} = 0$

On the other hand,

$\lim_{x \to 0^-}e^{\frac{1}{x}} = 0$ implies $\lim_{x \to 0^-} \frac{1}{1+e^{\frac{1}{x}}} = 1$

So this limit does not exist and, in particular, is different from $f(0) = 0$.

Therefore the function is not continuous at $x=0$