What is the limit as h approaches 0 of (ln(2e+h)-ln(2e))/h ?

The limit as h approaches 0 of (ln(2e+h)-ln(2e))/h is 1/(2e)

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limit as h approaches 0 of (ln(2e+h)-ln(2e))/h

h \to 0 lim (\frac{ln ( 2 e + h ) - ln ( 2 e )}{h})

\frac{1}{2 e}

Decimal Notation

0.18393 \dots

Solution steps

h \to 0 lim (\frac{ln ( 2 e + h ) - ln ( 2 e )}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{\frac{1}{h + 2 e}}{1})

= h \to 0 lim (\frac{\frac{1}{h + 2 e}}{1})

Plug in the value

h = 0

= \frac{\frac{1}{0 + 2 e}}{1}

Simplify

= \frac{1}{2 e}

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

x \to 0 lim (\frac{7 x - 1}{5 x})

x \to 1 lim (2 x - 4)

x \to 0 lim (\frac{x ^{2} - x}{∣ x ∣})

x \to 0 + lim (\frac{1}{tan ( x )})

What is the limit as h approaches 0 of (ln(2e+h)-ln(2e))/h ?
The limit as h approaches 0 of (ln(2e+h)-ln(2e))/h is 1/(2e)