What is the limit as h approaches 0 of ((2+h)^{4+h}-16)/h ?

The limit as h approaches 0 of ((2+h)^{4+h}-16)/h is 16(2+ln(2))

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limit as h approaches 0 of ((2+h)^{4+h}-16)/h

h \to 0 lim (\frac{( 2 + h ) ^{4 + h} - 16}{h})

16 (2 + ln (2))

Decimal

43.09035 \dots

Solution steps

h \to 0 lim (\frac{( 2 + h ) ^{4 + h} - 16}{h})

Apply L'Hopital's Rule

= h \to 0 lim (\frac{( \frac{h + 4}{h + 2} + ln ( h + 2 ) ) ( h + 2 ) ^{h + 4}}{1})

Plug in the value

h = 0

= \frac{( \frac{0 + 4}{0 + 2} + ln ( 0 + 2 ) ) ( 0 + 2 ) ^{0 + 4}}{1}

Simplify

\frac{( \frac{0 + 4}{0 + 2} + ln ( 0 + 2 ) ) ( 0 + 2 ) ^{0 + 4}}{1} : 16 (2 + ln (2))

= 16 (2 + ln (2))

What is the limit as h approaches 0 of ((2+h)^{4+h}-16)/h ?
The limit as h approaches 0 of ((2+h)^{4+h}-16)/h is 16(2+ln(2))