What is the best method to solve the limit limx→∞(1+sin⁡2x2)x2?

Hint:limx→∞(1+sin⁡2x2)x2=(limx→∞(1+sin⁡2x2)1sin⁡(2x2))2limx→∞sin⁡2x22x2Now for the exponent set 2x2=h⇒h→0+

For questions like this one, I've found asymptotic calculations to be handy. For example:(1+sin⁡2x2)x2=ex2ln⁡(1+sin⁡2x2)=ex2(sin⁡2x2+o(sin⁡2x2))=ex2(2x2+o(2x2)+o(2x2))=e2+o(1)→e2(x→∞)

Hint:limx→∞(1+sin⁡2x2)x2=(limx→∞(1+sin⁡2x2)1sin⁡(2x2))2limx→∞sin⁡2x22x2Now for the exponent set 2x2=h⇒h→0+

For questions like this one, I've found asymptotic calculations to be handy. For example:(1+sin⁡2x2)x2=ex2ln⁡(1+sin⁡2x2)=ex2(sin⁡2x2+o(sin⁡2x2))=ex2(2x2+o(2x2)+o(2x2))=e2+o(1)→e2(x→∞)

Resolved: "What is the best method to solve the..."

painter555ui8n

Answered question

2022-04-17

What is the best method to solve the limit

lim_{x \to \infty} {(1 + \sin \frac{2}{x^{2}})}^{x^{2}}

Answer & Explanation

madking711kzcb

Beginner2022-04-18Added 9 answers

Hint:

lim_{x \to \infty} {(1 + \sin \frac{2}{x^{2}})}^{x^{2}} = {(lim_{x \to \infty} {(1 + \sin \frac{2}{x^{2}})}^{\frac{1}{\sin (\frac{2}{x^{2}})}})}^{2 lim_{x \to \infty} \frac{\sin \frac{2}{x^{2}}}{\frac{2}{x^{2}}}}

Now for the exponent set

\frac{2}{x^{2}} = h \Rightarrow h \to 0^{+}

chabinka61jx

Beginner2022-04-19Added 12 answers

For questions like this one, I've found asymptotic calculations to be handy. For example:

{(1 + \sin \frac{2}{x^{2}})}^{x^{2}} = e^{x^{2} \ln (1 + \sin \frac{2}{x^{2}})} = e^{x^{2} (\sin \frac{2}{x^{2}} + o (\sin \frac{2}{x^{2}}))} = e^{x^{2} (\frac{2}{x^{2}} + o (\frac{2}{x^{2}}) + o (\frac{2}{x^{2}}))} = e^{2 + o (1)} \to e^{2} (x \to \infty)

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