What is the limit as t approaches 0 of (e^{6t}-1)/(sin(t)) ?

The limit as t approaches 0 of (e^{6t}-1)/(sin(t)) is 6

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limit as t approaches 0 of (e^{6t}-1)/(sin(t))

t \to 0 lim (\frac{e ^{6 t} - 1}{sin ( t )})

6

Solution steps

t \to 0 lim (\frac{e ^{6 t} - 1}{sin ( t )})

Apply L'Hopital's Rule

: t \to 0 lim (\frac{6 e ^{6 t}}{cos ( t )})

= t \to 0 lim (\frac{6 e ^{6 t}}{cos ( t )})

Plug in the value

t = 0

= \frac{6 e ^{6 \cdot 0}}{cos ( 0 )}

Simplify

\frac{6 e ^{6 \cdot 0}}{cos ( 0 )} : 6

= 6

θ \to + 0 lim (\frac{sin ( 8 θ )}{θ})

\int e^{- 9 x} d x

5 ln (x), x ln (x)

\int \frac{1}{3 x ^{2} - 2 x + 5} d x

\frac{d}{d x} (x^{4} - 8 x^{2} + 3)

What is the limit as t approaches 0 of (e^{6t}-1)/(sin(t)) ?
The limit as t approaches 0 of (e^{6t}-1)/(sin(t)) is 6