What is the limit as x approaches 0-of ln(2x^2+e) ?

The limit as x approaches 0-of ln(2x^2+e) is 1

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limit as x approaches 0-of ln(2x^2+e)

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

1

Solution steps

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

Plug in the value

x = 0

= ln (2 \cdot 0^{2} + e)

Simplify

ln (2 \cdot 0^{2} + e) : 1

= 1

x \to \infty lim (1 0^{x} + 2)

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

x \to 3 lim (2 x - 5)

x \to \frac{π}{6} lim (cot (4 x))

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

What is the limit as x approaches 0-of ln(2x^2+e) ?
The limit as x approaches 0-of ln(2x^2+e) is 1