Prove by using the formal definition of limit that:
$\displaystyle\lim_{x\to \infty }\dfrac{1}{x^4 + x^2 + 5} = 0$
Can anyone please lend a hand here?
Prove by using the formal definition of limit that:
$\displaystyle\lim_{x\to \infty }\dfrac{1}{x^4 + x^2 + 5} = 0$
Can anyone please lend a hand here?
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Hint: $$ \frac{1}{x^4+x^2+5} \le \frac{1}{x^2}\le \epsilon, $$ whenever $x > \sqrt{1/\epsilon}$.