What is the limit as x approaches 0 of (3^{sin(x)}-1)/(4x) ?

The limit as x approaches 0 of (3^{sin(x)}-1)/(4x) is 1/4 ln(3)

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limit as x approaches 0 of (3^{sin(x)}-1)/(4x)

x \to 0 lim (\frac{3 ^{s i n (x)} - 1}{4 x})

\frac{1}{4} ln (3)

Decimal Notation

0.27465 \dots

Solution steps

x \to 0 lim (\frac{3 ^{s i n (x)} - 1}{4 x})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{4} \cdot x \to 0 lim (\frac{3 ^{s i n (x)} - 1}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{ln ( 3 ) \cdot 3 ^{s i n (x)} cos ( x )}{1})

= \frac{1}{4} \cdot x \to 0 lim (\frac{ln ( 3 ) \cdot 3 ^{s i n (x)} cos ( x )}{1})

Plug in the value

x = 0

= \frac{1}{4} \cdot \frac{ln ( 3 ) \cdot 3 ^{s i n (0)} cos ( 0 )}{1}

Simplify

\frac{1}{4} \cdot \frac{ln ( 3 ) \cdot 3 ^{s i n (0)} cos ( 0 )}{1} : \frac{1}{4} ln (3)

= \frac{1}{4} ln (3)

25 y^{^{''}} + 10 y = 0

\frac{d}{d x} (- ln (4 x))

\int (- 8 sec (x) tan (x) - 3 sec^{2} (x)) d x

y^{^{''}} + 5 y^{^{'}} - 6 y = (3 x - 1) e^{x}

e^{x^{4}}

What is the limit as x approaches 0 of (3^{sin(x)}-1)/(4x) ?
The limit as x approaches 0 of (3^{sin(x)}-1)/(4x) is 1/4 ln(3)