Compute $\lim_{n\rightarrow \infty} f_n(x)$, determine if f(x) is continuous and show that $f_n(x)$ does not converge uniformly on R for the sequence of functions $f_n(x) = \frac{x^2}{x^2+(1+nx)^2}$
attempt:
$\large \lim_{n\rightarrow \infty} f_n(x) = \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)} $
$\large = \frac{\lim_{n\rightarrow \infty}x^2}{\lim_{n\rightarrow \infty}x^2+\lim_{n\rightarrow \infty}(1+nx)^2}$
$\large = \frac{x^2}{x^2+ \infty} = \frac{x^2}{\infty} = 0$
$f_n(x)$ does not converge uniformly if $\exists \epsilon>0: \forall N \exists n > N \exists x=-\frac{1}{n} \epsilon R \implies |f_n(x)-f(x)| = |f_n(-\frac{1}{n})-0|$ $\large = |\frac{\frac{1}{n^2}}{\frac{1}{n^2}+(1+n(\frac{-1}{n}))^2}-0|$ $\large =|\frac{\frac{1}{n^2}}{\frac{1}{n^2}}-0| = |1-0| = 1 \geq \epsilon = 1$ So $f_n(x)$ does not converge uniformly on R.
Also, f(x) is not continuous because $f_n(x)$ does not converge to f(x) uniformly on R.