Let [math: f(x)=x2sin⁡(x)]f(x) = x^2 \sin(x) and let [math: xn=1/n]x_n=1/n. What is the limit of the sequence [math: {f(xn)}]\{f(x_n)\}?

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BlackTom AI · Accepted Answer

We are given f(x) = x^2 sin(x) and x_n = 1/n, and we want the limit of f(x_n) as n → ∞.

First, substitute x_n into f: f(x_n) = (1/n)^2 · sin(1/n).

For small values of t, sin(t) ~ t. Here t = 1/n, which tends to 0 as n grows. Therefore sin(1/n) behaves like 1/n for large n, and we can approximate:
 f(x_n) ≈ (1/n)^2 · (1/n) = 1/n^3.

More rigorously, we can use the limit sin(t) / t → 1 as t → 0, which implies sin(1/n) = (1/n) · (sin(1/n) / (1/n)) → (1/n) · 1 = 1/n. Substituting back gives f(x_n) = (1/n)^2 · sin(1/n) = (1/n)^2 · (1/n) · (sin(1/n)/(1/n)) → (1/n)^3 · 1, which again tends to 0 as n → ∞.

Therefore, the sequence f(x_n) behaves like 1/n^3 for large n and its limit is 0.

Let [math: f(x)=x2sin⁡(x)]f(x) = x^2 \sin(x) and let [math: xn=1/n]x_n=1/n. What is the limit of the sequence [math: {f(xn)}]\{f(x_n)\}?

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Let [math: f(x)=x2sin⁡(x)]f(x) = x^2 \sin(x) and let [math: xn=1/n]x_n=1/n. What is the limit of the sequence [math: {f(xn)}]\{f(x_n)\}?

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Similar Questions

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If the positive-value sequences \(\{x_n\}, \{y_n\}\) have limits 5 and 3 respectively, what are the limits of the sequences \(\{x_n+y_{n+1}\}\), \(\{x_ny_{2n}\}\), \(\{2x_n\}\) and \(\{x_n^{y_n}\} \).Enter your answer in the form *,*,*,*

Let [math: f(x)=x2]f(x) = x^2 and [math: xn=2+1/n]x_n=2+1/n.What is the limit of the sequence [math: {f(xn)}]\{f(x_n)\}?

If the positive-value sequences [math: {xn},{yn}]\{x_n\}, \{y_n\} have limits 5 and 3 respectively, what are the limits of the sequences [math: {xn+yn+1}]\{x_n+y_{n+1}\}, [math: {xny2n}]\{x_ny_{2n}\}, [math: {2xn}]\{2x_n\} and [math: {xnyn}]\{x_n^{y_n}\} .Enter your answer in the form *,*,*,*

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