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

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

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

x \to 0 lim (\frac{7 ^{s i n (x)} - 1}{x})

ln (7)

Decimal

1.94591 \dots

Solution steps

x \to 0 lim (\frac{7 ^{s i n (x)} - 1}{x})

Apply L'Hopital's Rule

= x \to 0 lim (\frac{ln ( 7 ) \cdot 7 ^{s i n (x)} cos ( x )}{1})

Plug in the value

x = 0

= \frac{ln ( 7 ) \cdot 7 ^{s i n (0)} cos ( 0 )}{1}

Simplify

\frac{ln ( 7 ) \cdot 7 ^{s i n (0)} cos ( 0 )}{1} : ln (7)

= ln (7)

\int \frac{1}{9} x^{- \frac{8}{9}} d x

\frac{\partial}{\partial x} (x^{3} e^{y} sin (x - y))

t an g e n t f (x) = x^{2} - 3, (3, 6)

ma c l a u r in \frac{4 x}{x ^{2} + 2 x - 3}

\frac{d}{d x} (\frac{1}{5 - 11 e ^{x}})

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