The value of lim x- 0 1- 5x 2 1- 3x 2 1 x 2 is

The correct option is B e^2 lim_x → 0( 1 + 5x^21 + 3x^2)^1x^2 = lim_x → 0( 1 + 5x^2)^1x^2lim_x → 0( 1 + 3x^2)^1x^2 = e^5x^2×1x^2e^3x^2×1x^2 ...

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Substitution Method to Remove Indeterminate Form

The value of ...

Question

The value of $lim x \to 0 {(\frac{1 + {5 x}^{2}}{1 + {3 x}^{2}})}^{\frac{1}{x^{2}}}$ is

e^{2}

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e

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e^{3}

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e^{4}

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lim x \to 0 {(\frac{1 + 5 x^{2}}{1 + 3 x^{2}})}^{1 / x^{2}} =

lim x \to 0 {(\frac{1 + 5 x^{2}}{1 + 3 x^{2}})}^{1 / x^{2}}

e = 2.72, e^{2} = 7.39, e^{3} = 20.09

e^{4} = 54.60

\int_{0}^{4} e^{x} d x

h = 1

E = E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}

P (E_{1}) = \frac{95}{100}, P (E_{2} / E_{1}) = \frac{94}{99}, P (E_{3} / (E_{1} \cap E_{2})) = \frac{93}{98}, P (E_{4} / E_{1} \cap E_{2} \cap E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) = \frac{91}{96}

P (E)

E = E_{1} E_{2} E_{3} E_{4} E_{5}

P (E_{1}) = \frac{95}{100}

P (E_{2} / E_{1}) = \frac{94}{99}

P (E_{1} / E_{1} E_{2}) = \frac{93}{98}

P (E_{4} / E_{1} E_{2} E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} E_{2} E_{3} E_{4}) = \frac{91}{96}

P (E)

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