What is the limit as x approaches 0 of (ln(x^2+1))/(x+4*sin(x)) ?

The limit as x approaches 0 of (ln(x^2+1))/(x+4*sin(x)) is 0

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limit as x approaches 0 of (ln(x^2+1))/(x+4*sin(x))

x \to 0 lim (\frac{ln ( x ^{2} + 1 )}{x + 4 \cdot sin ( x )})

0

Solution steps

x \to 0 lim (\frac{ln ( x ^{2} + 1 )}{x + 4 sin ( x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{\frac{2 x}{x ^{2} + 1}}{1 + 4 cos ( x )})

= x \to 0 lim (\frac{\frac{2 x}{x ^{2} + 1}}{1 + 4 cos ( x )})

Plug in the value

x = 0

= \frac{\frac{2 \cdot 0}{0 ^{2} + 1}}{1 + 4 cos ( 0 )}

Simplify

\frac{\frac{2 \cdot 0}{0 ^{2} + 1}}{1 + 4 cos ( 0 )} : 0

= 0

x \to 0 lim (x^{3} \cdot cos (\frac{1}{x}))

x \to 0 lim (\frac{e ^{x}}{1})

n \to \infty lim (n)

x \to \infty lim (- 1^{x} \cdot x)

x \to \infty lim (\frac{100 x ^{4} - 5 x ^{3} + 1}{2 - 2 x ^{2} - 10 x ^{4}})

What is the limit as x approaches 0 of (ln(x^2+1))/(x+4*sin(x)) ?
The limit as x approaches 0 of (ln(x^2+1))/(x+4*sin(x)) is 0