lim-x-0-1-x-2-x-1-x-3-x-2-x-2-

Question Number 140000 by bramlexs22 last updated on 03/May/21

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

Commented by MJS_new last updated on 03/May/21

the answer should be \frac{4}{9} I think

Commented by bramlexs22 last updated on 03/May/21

yes the answer is \frac{4}{9}

Answered by TheSupreme last updated on 03/May/21

{(\frac{1 + l n (2) x^{2}}{1 + l n (3) x^{2}})}^{\frac{2}{x^{2}}}

u = \frac{1}{x^{2}}

l i m {(\frac{1 + \frac{l n (2)}{u^{2}}}{1 + \frac{l n (3)}{u^{2}}})}^{2 u^{2}} = l i m {(\frac{u^{2} + l n (2)}{u^{2} + l n (3)})}^{2 u^{2}} = l i m {(1 + \frac{l n (2) - l n (3)}{l n (3) + u^{2}})}^{2 u^{2}}

v = u^{2} + l n (3)

l i m {(1 + \frac{l n (2) - l n (3)}{v})}^{2 v - 2 l n (3)}

{[e^{\frac{2}{l n (2) - l n (3)}}]}^{}

Answered by bobhans last updated on 03/May/21

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

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

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

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

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

\ln L = \ln (\frac{4}{9}) \Rightarrow L = e^{\ln (\frac{4}{9})}

L = \frac{4}{9}

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