Compute the limit of the sequence of functions of $\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}$ Compute the limit of the sequence of functions of 
$$
\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}.
$$
Attempt.
$$\lim_{n\to \infty} f_n(x) = \lim_{n\to \infty} \frac{x^2}{x^2+(1-nx)^2} = \lim_{n\to \infty} \frac{x^2}{x^2+n^2x^2-2nx + 1} $$
$$= \lim_{n\to \infty} \frac{x^2}{x^2+\frac{x^2}{n^2}-\frac{2x}{n} + \frac{1}{n}} = \lim_{n\to \infty} \frac{x^2}{x^2} = 1$$
 A: Well $$\lim \frac{x^2}{x^2+n^2x^2-2nx + 1} \ne  \lim \frac{x^2}{x^2+\frac{x^2}{n^2}-\frac{2x}{n} + \frac{1}{n}} $$
Note that there is no need to do any algebraic manipulation here; you can just do it by inspection as $n \to \infty$ we immediately see that $$\lim_{n\to \infty}  \frac{x^2}{x^2+(1-nx)^2}=0$$
as the denominator is tends to infinity the fraction must be zero.
A: For $x=0$ and all $n\in \mathbf{N}$, we see that $$f_{n}(0)=\frac{0}{0+1}=0$$ For $x\neq 0$, then we note that $$\lim_{n\rightarrow \infty}(1-nx)$$ is $+\infty$ for $x<0$ and is $-\infty$ for $x>0$.  Thus, $$\lim_{n\rightarrow \infty}(1-nx)^{2}=\left(\lim_{n\rightarrow \infty}(1-nx)\right)^{2}=+\infty$$ and $$\lim_{n\rightarrow \infty} (x^{2}+(1-nx)^{2})=x^{2}+\lim_{n\rightarrow \infty}(1-nx)^{2}=x^{2}+\infty=\infty$$ (remember that while we take the sequential limit, we are holding $x$ fixed).  It follows that $$\lim_{n\rightarrow \infty}\frac{x^{2}}{x^{2}+(1-nx)^{2}}=\frac{\lim_{n\rightarrow \infty}x^{2}}{\lim_{n\rightarrow \infty}(x^{2}+(1-nx)^{2})}=\frac{x^{2}}{\infty}=0$$ since we have restricted our attentions to $x\neq 0$.  Therefore, $$\lim_{n\rightarrow \infty}f_{n}(x)=0$$ for all real $x$.
