CALC 2.4 QNO 6

Given the function: $f(x) = \frac{8}{1 + e^{\frac{1}{x}}}$

(a) When x gets close to 0 from the left side (negative side), the term $e^{\frac{1}{x}}$ gets really close to 0. So, our function is almost like: $f(x) = \frac{8}{1}$ Which is 8.

So: $\lim_{x \to 0^-} f(x) = 8$

(b) When x gets close to 0 from the right side (positive side), the term $e^{\frac{1}{x}}$ becomes really huge. Because of this, our function is almost like dividing 8 by a super big number, which gives us something very close to 0.

So: $\lim_{x \to 0^+} f(x) = 0$

(c) When trying to see what happens exactly at 0, we notice that the value from the left side (8) is not the same as the value from the right side (0). So, we can't say exactly what the function is at 0.

Therefore: $\lim_{x \to 0} f(x) = \text{DNE}$