What is the limit as x approaches 0 of 4/(3+e^{1/x)} ?

The limit as x approaches 0 of 4/(3+e^{1/x)} is diverges

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limit as x approaches 0 of 4/(3+e^{1/x)}

x \to 0 lim (\frac{4}{3 + e ^{\frac{1}{x}}})

diverges

Solution steps

x \to 0 lim (\frac{4}{3 + 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{4}{3 + e ^{\frac{1}{x}}}) = 0

x \to 0 - lim (\frac{4}{3 + e ^{\frac{1}{x}}}) = \frac{4}{3}

= diverges

\frac{d}{d x} (sin (2 x + 3 y))

\frac{\partial}{\partial y} (ln (x^{2} + y))

x^{4} - 2 x^{2}, 2 x^{2}, [0, 2]

f (x) = \frac{8 x}{x + 7}, (2, 1.7777700000000 \dots)

y = 5^{x - 3}, (3, 1)

What is the limit as x approaches 0 of 4/(3+e^{1/x)} ?
The limit as x approaches 0 of 4/(3+e^{1/x)} is diverges