Could someone please check if I have the correct solution for this please.
I need to find the limit as n tends to infinity of $$ f_n(x) = \frac{1}{n^2(1+x^{2n})} $$
For $x=0$ we have
$f(0)=1/0$ which is undefined, where do I go from here...
Could someone please check if I have the correct solution for this please.
I need to find the limit as n tends to infinity of $$ f_n(x) = \frac{1}{n^2(1+x^{2n})} $$
For $x=0$ we have
$f(0)=1/0$ which is undefined, where do I go from here...
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$0 \leq f _n(x) \leq \frac 1 {n^{2}}$. By Squeeze Theorem $\lim_{n\to \infty} f_n(x)=0$ for any real number $x$.