What is the limit as x approaches 0 of 4x-ln(4x+1) ?

The limit as x approaches 0 of 4x-ln(4x+1) is 0

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limit as x approaches 0 of 4x-ln(4x+1)

x \to 0 lim (4 x - ln (4 x + 1))

0

Solution steps

x \to 0 lim (4 x - ln (4 x + 1))

Plug in the value

x = 0

= 4 \cdot 0 - ln (4 \cdot 0 + 1)

Simplify

4 \cdot 0 - ln (4 \cdot 0 + 1) : 0

= 0

x \to 0 lim (\frac{12}{x ^{2}})

y^{^{''}} + y^{^{'}} + a^{2} y = 0

\int \frac{1 + 6 x}{( 4 x - 3 ) ( 2 x + 5 )} d x

5 x y^{^{'}} = 2 x e^{x} - 5 y + 6 x^{2}

x \to 0 + lim (\frac{7 x}{x})

What is the limit as x approaches 0 of 4x-ln(4x+1) ?
The limit as x approaches 0 of 4x-ln(4x+1) is 0