What is the limit as h approaches 0 of ((1+h)^{3pi}-1)/h ?

The limit as h approaches 0 of ((1+h)^{3pi}-1)/h is 3pi

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limit as h approaches 0 of ((1+h)^{3pi}-1)/h

h \to 0 lim (\frac{( 1 + h ) ^{3 π} - 1}{h})

3 π

Decimal

9.42477 \dots

Solution steps

h \to 0 lim (\frac{( 1 + h ) ^{3 π} - 1}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{3 π ( h + 1 ) ^{3 π - 1}}{1})

= h \to 0 lim (\frac{3 π ( h + 1 ) ^{3 π - 1}}{1})

Plug in the value

h = 0

= \frac{3 π ( 0 + 1 ) ^{3 π - 1}}{1}

Simplify

= 3 π

x \to - \infty lim (- 17 x^{2})

x \to 100 lim (x)

x \to - 4 lim (x^{2} - 10)

x \to \frac{π}{2} - lim (sin (x))

m \to 0 lim (\frac{m}{\frac{1}{m}})

What is the limit as h approaches 0 of ((1+h)^{3pi}-1)/h ?
The limit as h approaches 0 of ((1+h)^{3pi}-1)/h is 3pi