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

Question Number 160569 by LEKOUMA last updated on 02/Dec/21

C a l c u l a t e

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

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

Answered by qaz last updated on 02/Dec/21

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

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

= \lim_{x \to 0} {[{(1 + \frac{2 x}{x + 1} + o (\frac{x}{x + 1}))}^{\frac{x + 1}{2 x}}]}^{\frac{2 (x^{2} + 1)}{x + 1}}

= e^{2}

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

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

= e \lim_{x \to 0, ξ \to 1} \frac{1}{ξ x^{2}} [(1 + 2^{x} x) - (1 + 3^{x} x)]

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

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

= \frac{2}{3}

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