Calculate-lim-x-0-1-x-2-x-1-x-3-x-1-x-2-lim-x-0-2e-x-x-1-1-x-2-1-x-lim-x-a-x-x-a-a-x-a-

Question Number 161723 by LEKOUMA last updated on 21/Dec/21

C a l c u l a t e

\lim_{x \to 0} {(\frac{1 + x . 2^{x}}{1 + x . 3^{x}})}^{\frac{1}{x^{2}}}

\lim_{x \to 0} {[2 e^{\frac{x}{x + 1}} - 1]}^{\frac{x^{2} + 1}{x}}

\lim_{x \to a} \frac{x^{x} - a^{a}}{x - a}

Answered by cortano last updated on 21/Dec/21

(1) \lim_{x \to 0} {(\frac{1 + x . 2^{x}}{1 + x . 3^{x}})}^{\frac{1}{x^{2}}} = e^{\lim_{x \to 0} \frac{x (2^{x} - 3^{x})}{x^{2}}}

= e^{\lim_{x \to 0} (\frac{2^{x} - 3^{x}}{x})} = e^{\ln 2 - \ln 3} = \frac{2}{3}

Answered by cortano last updated on 21/Dec/21

(2) \lim_{x \to 0} {(2 e^{\frac{x}{x + 1}} - 1)}^{\frac{x^{2} + 1}{x}} = e^{\lim_{x \to 0} \frac{2 (e^{\frac{x}{x + 1}} - 1) (x^{2} + 1)}{x}}

= e^{\underset{x \to 0}{\lim 2} (\frac{1}{{(x + 1)}^{2}} . e^{\frac{x}{x + 1}})} = e^{2}

Commented by LEKOUMA last updated on 21/Dec/21

M a n y t h a n k s

Answered by Ar Brandon last updated on 21/Dec/21

A = \lim_{x \to 0} {(\frac{1 + x \cdot 2^{x}}{1 + x . 3^{x}})}^{\frac{1}{x^{2}}} \Rightarrow \ln A = \lim_{x \to 0} \frac{1}{x^{2}} \ln (\frac{1 + x . 2^{x}}{1 + x . 3^{x}})

\ln A = \lim_{x \to 0} \frac{1}{x^{2}} (x . 2^{x} - x . 3^{x}) = \lim_{x \to 0} \frac{2^{x} - 3^{x}}{x} = \ln (\frac{2}{3})

\ln A = \ln (\frac{2}{3}) \Rightarrow \begin{matrix} A = \frac{2}{3} \end{matrix}

Answered by Ar Brandon last updated on 21/Dec/21

B = \lim_{x \to a} \frac{x^{x} - a^{a}}{x - a} = \lim_{x \to a} x^{x} (1 + \ln x) = a^{a} (1 + \ln a)

Commented by LEKOUMA last updated on 21/Dec/21

M a n y t h a n k s

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