Limx - x2 - 5x - 3x2 - x - 3x -

The correct option is B e^4 lim_x→∞(x^2+5x+3x^2+x+3)^x #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; # ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sum of n Terms

limx →∞ x2 + ...

Question

$lim x \to \infty {(\frac{x^{2} + 5 x + 3}{x^{2} + x + 3})}^{x} =$

e^{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

e^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

lim x \to \infty {(\frac{x^{2} + 5 x + 3}{x^{2} + x + 2})}^{x} = e^{4}

lim x \to \infty (1 + x)^{1 / x} = e

lim x \to \infty {(\frac{x^{2} + 5 x + 3}{x^{2} + x + 2})}^{x}

E_{1}, E_{2}, E_{3} a n d E_{4}

1^{s t}, 2^{n d}, 3^{r d}

4^{t h}

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)

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)

Join BYJU'S Learning Program