Question

Please help,how do you find the limit of ((x^2 + 1)/(x^2 - 2))^x^2,
when x approaches infinity?

Answer 1

\[\ln((\frac{ x ^{2}+1 }{ x ^{2}-2 })^{x ^{2}})=x ^{2}\ln(\frac{ x ^{2}+1 }{ x ^{2}-2 })=\frac{ \ln(\frac{ x ^{2}+1 }{ x ^{2}-2 }) }{ \frac{ 1 }{ x ^{2} } }\] Let's put : \[f(x)=\ln(\frac{ x ^{2}+1 }{ x ^{2}-2 })\] \[g(x)=\frac{ 1 }{ x ^{2} }\] lim f(x) =lim g(x) = 0. From l'Hôpital's rule we get: \[\lim_{x \rightarrow +\infty}\frac{ f(x) }{ g(x) }=\lim_{x \rightarrow +\infty}\frac{ f'(x) }{ g'(x) }=\lim_{x \rightarrow +\infty}\frac{ -6x }{ 2x(x ^{2}+1)(x ^{2}-2) }=0\] So: \[\lim_{x \rightarrow +\infty}(\frac{ x ^{2}+1 }{ x ^{2}-2 })^{x ^{2}}=e ^{0}=1\]

Answer 2

According to Wolfram the correct answer is: \[e^{3}\]

Answer 3

(x^2 + 1)/(x^2 - 2) = 1 + 3/(x^2 - 1)

\[ \lim_{x\to \infty} \left( 1 + \frac 1 {x^2 - 2}\right)^{x^2} = e^3\]

Answer 4

*
\[ \lim_{x\to \infty} \left( 1 + \frac 3 {x^2 - 2}\right)^{x^2} = e^3 \]

Answer 5

use the definition of
\[ \lim_{x\to \infty} (1 + a/x)^x = e^a\]

Answer 6

also x->infty => x^2 - 2 -> infty
\[\lim_{x\to \infty} \left( 1 + \frac 3 {x^2 - 2}\right)^{x^2} = \lim_{x^2 - 2\to \infty} \left( 1 + \frac 3 {x^2 - 2}\right)^{x^2-2} *\lim_{x\to \infty} \left( 1 + \frac 3 {x^2 - 2}\right)^{2} \]

Answer 7

Yeah, made a little mistake there in the derivative. \[\lim_{x \rightarrow +\infty}\frac{ f'(x) }{ g'(x) }= \lim_{x \rightarrow +\infty}\frac{ 3x ^{4} }{ (x ^{2}+1)(x ^{2}-2) }=3\] So: \[\lim_{x \rightarrow +\infty}(\frac{ x ^{2}+1 }{ x ^{2}-2 })^{x ^{2}}=e ^{3}\]