What is the limit as x approaches 0 of 8/(2+e^{1/x)} ?

The limit as x approaches 0 of 8/(2+e^{1/x)} is diverges

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limit as x approaches 0 of 8/(2+e^{1/x)}

x \to 0 lim (\frac{8}{2 + e ^{\frac{1}{x}}})

diverges

Solution steps

x \to 0 lim (\frac{8}{2 + e ^{\frac{1}{x}}})

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

x \to 0 + lim (\frac{8}{2 + e ^{\frac{1}{x}}}) = 0

x \to 0 - lim (\frac{8}{2 + e ^{\frac{1}{x}}}) = 4

= diverges

\frac{d}{d x} (\frac{2 cos ( x )}{1 + sin ( x )})

x \to \infty lim (x^{2} - x)

f (x) = \frac{12}{x}

\frac{d}{d x} (sec^{3} (x))

n = 1 \sum \infty (ln (3))^{n}

What is the limit as x approaches 0 of 8/(2+e^{1/x)} ?
The limit as x approaches 0 of 8/(2+e^{1/x)} is diverges