What is the limit as x approaches 0 of e^{1/x ln(1+x^2)} ?

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

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limit as x approaches 0 of e^{1/x ln(1+x^2)}

x \to 0 lim (e^{\frac{1}{x} l n (1 + x^{2})})

1

Solution steps

x \to 0 lim (e^{\frac{1}{x} l n (1 + x^{2})})

Apply the Limit Chain Rule:

1

= 1

\frac{\partial}{\partial x} (ln (\frac{x}{x + 2}))

\int \frac{3 x ^{2} - 2}{x ^{2} - 2 x - 8} d x

\int (y - 4)^{2} d y

\frac{d}{d x} (\frac{x ^{2}}{x - 6})

\frac{\partial}{\partial x} (\frac{x y}{1 + x y ^{2}})

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