What is the limit as x approaches 1 of (ln(x))/(4x-x^2-3) ?

The limit as x approaches 1 of (ln(x))/(4x-x^2-3) is 1/2

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limit as x approaches 1 of (ln(x))/(4x-x^2-3)

x \to 1 lim (\frac{ln ( x )}{4 x - x ^{2} - 3})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

x \to 1 lim (\frac{ln ( x )}{4 x - x ^{2} - 3})

Apply L'Hopital's Rule

: x \to 1 lim (\frac{\frac{1}{x}}{- 2 x + 4})

= x \to 1 lim (\frac{\frac{1}{x}}{- 2 x + 4})

Plug in the value

x = 1

= \frac{\frac{1}{1}}{- 2 \cdot 1 + 4}

Simplify

= \frac{1}{2}

\frac{d}{d x} ((5 - x^{2})^{5})

x \to 0 + lim (x^{- 2})

x \to 2 lim (ln ∣ 2 - x ∣)

\int 5 x^{2} e^{x^{3}} d x

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

What is the limit as x approaches 1 of (ln(x))/(4x-x^2-3) ?
The limit as x approaches 1 of (ln(x))/(4x-x^2-3) is 1/2