What is the limit as t approaches 0 of (e^{7t}-1)/(sin(t)) ?

The limit as t approaches 0 of (e^{7t}-1)/(sin(t)) is 7

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limit as t approaches 0 of (e^{7t}-1)/(sin(t))

t \to 0 lim (\frac{e ^{7 t} - 1}{sin ( t )})

7

Solution steps

t \to 0 lim (\frac{e ^{7 t} - 1}{sin ( t )})

Apply L'Hopital's Rule

: t \to 0 lim (\frac{7 e ^{7 t}}{cos ( t )})

= t \to 0 lim (\frac{7 e ^{7 t}}{cos ( t )})

Plug in the value

t = 0

= \frac{7 e ^{7 \cdot 0}}{cos ( 0 )}

Simplify

\frac{7 e ^{7 \cdot 0}}{cos ( 0 )} : 7

= 7

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What is the limit as t approaches 0 of (e^{7t}-1)/(sin(t)) ?
The limit as t approaches 0 of (e^{7t}-1)/(sin(t)) is 7