What is the limit as x approaches 0 of (sin(3x^2))/(ln(x^2+1)) ?

The limit as x approaches 0 of (sin(3x^2))/(ln(x^2+1)) is 3

Popular >

limit as x approaches 0 of (sin(3x^2))/(ln(x^2+1))

x \to 0 lim (\frac{sin ( 3 x ^{2} )}{ln ( x ^{2} + 1 )})

3

Solution steps

x \to 0 lim (\frac{sin ( 3 x ^{2} )}{ln ( x ^{2} + 1 )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{cos ( 3 x ^{2} ) \cdot 6 x}{\frac{2 x}{x ^{2} + 1}})

= x \to 0 lim (\frac{cos ( 3 x ^{2} ) \cdot 6 x}{\frac{2 x}{x ^{2} + 1}})

Simplify

\frac{cos ( 3 x ^{2} ) \cdot 6 x}{\frac{2 x}{x ^{2} + 1}} : 3 cos (3 x^{2}) (x^{2} + 1)

= x \to 0 lim (3 cos (3 x^{2}) (x^{2} + 1))

Plug in the value

x = 0

= 3 cos (3 \cdot 0^{2}) (0^{2} + 1)

Simplify

3 cos (3 \cdot 0^{2}) (0^{2} + 1) : 3

= 3

\frac{d}{d x} (\frac{x}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}})

e^{t a n (θ)}

\frac{\partial}{\partial x} (y ln (3 x + 5 y))

x, 2 x - 15, 0

What is the limit as x approaches 0 of (sin(3x^2))/(ln(x^2+1)) ?
The limit as x approaches 0 of (sin(3x^2))/(ln(x^2+1)) is 3