Question

Limit of x tending 0 e power x minus x minus one upon x

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Please answer

Answer 1

we should find \( \lim _{x \to 0}(e^x-x-\frac{1}{x}) \).

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

we know \( \lim _{x \to 0}(e^x)=e^0=1 \) and \( \lim _{x \to 0}(x)=0 \) and \( \lim _{x \to 0^+}(\frac{1}{x})=+ \infty \) and \( \lim _{x \to 0^-}(\frac{1}{x})=- \infty \). so:

\( \lim _{x \to 0}(e^x-x-\frac{1}{x})=1-0 \pm \infty=\pm \infty \)

Answer 2

\(\underset{x\to 0}{lim}\) e^x-1/x = e^0-1/0 (By taking limit)

= e^-∞ (∵ 1/0 = ∞)

\(=\frac{1}{e^\infty}\)

= 1/∞ (∵ e^∞ = ∞)

= 0 (∵ 1/∞ = 0)