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

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

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

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

diverges

Solution steps

x \to - 0 lim (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 (e^{\frac{1}{x}}) = \infty

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

= diverges

\frac{d}{d t} (5 cos (t))

\int_{0}^{1} \frac{1}{x ^{2} + 16} d x

y^{^{''}} - 5 y^{^{'}} - 2 y = 0

\int_{- 8}^{8} (8 - ∣ x ∣) d x

h \to 0 lim ((x + h)^{2} - x^{2})

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