A problem by Huzaifa

$\because$ The denominator has a higher degree than the numerator.

$\therefore$ As $x \to \infty$ , $\boxed{f(x) \to 0}$ , where $f(x) = \dfrac{100}{x^{2}+5}$ .

To solve the limit problems, if $\displaystyle \lim_{x \to 1} \frac{9x^{2} - 5x}{4x}$ , we can substitute $x = 1$ to the fraction, but if $\displaystyle \lim_{x \to \infty} \frac{5x^{2} - 6}{2x}$ , we cannot substitute $x = \infty$ to the fraction. But we can solve it.

If $\displaystyle \lim_{ x \to \infty } a$ , then the value of $a$ is 0, but it is not the absolute value.

Though $\displaystyle \lim_{ x \to \infty } a$ , we can find the value of $a$ is not $0$ .

If the value of $\displaystyle \lim_{ x \to \infty } a$ is not $0$ , then the value is $\infty$ .

But, $\displaystyle \lim_{ x \to \infty } \frac { 100 } { x^{ 2 } + 5 }$ , the value is $\boxed { 0 }$ .