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)\}?

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BlackTom AI · Accepted Answer

We start by identifying the given functions: f(x) = x^2 and the sequence x_n = 2 + 1/n. 
Next, we evaluate f at x_n by substitution: f(x_n) = (x_n)^2 = (2 + 1/n)^2. 
Expanding the square yields (2 + 1/n)^2 = 4 + 4/n + (1/n)^2. 
Now consider the limit as n → ∞: each term involving 1/n or 1/n^2 tends to 0, so 4 + 4/n + 1/n^2 approaches 4. 
Since 4/n → 0 and 1/n^2 → 0 as n grows without bound, the entire expression converges to 4. 
Intuitively, the sequence x_n approaches 2, and since f is continuous (being a polynomial), the limit of f(x_n) equals f(lim x_n) = f(2) = 4. 
Any deviation in the reasoning would require a different type of function or a non-convergent x_n, but given the current definitions, the convergence behavior is determined by the small perturbations 1/n around 2, which vanish in the limit.

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)\}?

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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)\}?

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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)\}?

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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