The limit of $2$ plus $3$ divided by $x$ square should be evaluated as the value of $x$ approaches infinity in this limit question. So, let’s learn how to find the limit of two plus three divided by square of $x$ as the value of $x$ tends to infinity.

The given algebraic function is formed by the addition of a constant function $2$ and an algebraic function $3$ divided by square of $x$. The limit of sum of the two functions $2$ and $3$ divided by $x$ square can be evaluated by finding the limits of them as per the sum rule of the limits.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize 2}$ $+$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(\dfrac{3}{x^2}\bigg)}$

Let’s first evaluate the limit of constant function $2$ as the value of $x$ approaches infinity by using the limit rule of a constant.

$=\,\,$ $2$ $+$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(\dfrac{3}{x^2}\bigg)}$

Now, it is time to focus on finding the limit of the $3$ divided by $x$ square. Firstly, let’s try to split the rational function as a product of two fractions.

$=\,\,$ $2$ $+$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(\dfrac{3 \times 1}{x^2}\bigg)}$

$=\,\,$ $2$ $+$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(3 \times \dfrac{1}{x^2}\bigg)}$

The number $3$ is a coefficient of the reciprocal of $x$ square. So, it can be released from the limit operation as per the constant multiple rule of the limits.

$=\,\,$ $2$ $+$ $3 \times \displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(\dfrac{1}{x^2}\bigg)}$

The reciprocal of $x$ square can be written as the square of the reciprocal of variable $x$ by the power of a quotient rule.

$=\,\,$ $2$ $+$ $3 \times \displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(\dfrac{1 \times 1}{x^2}\bigg)}$

$=\,\,$ $2$ $+$ $3 \times \displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(\dfrac{1^2}{x^2}\bigg)}$

$=\,\,$ $2$ $+$ $3 \times \displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(\dfrac{1}{x}\bigg)^2}$

Now, the power rule of limits should be used to find the limit of square of the multiplicative inverse of $x$ by the square of the limit of the reciprocal of $x$.

$=\,\,$ $2$ $+$ $3 \times \bigg(\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \dfrac{1}{x}\bigg)^2}$

According to the limit rule at infinity, the limit of the reciprocal of $x$ as the $x$ approaches infinity is equal to zero.

$=\,\,$ $2$ $+$ $3 \times (0)^2$

The limit of $2$ plus $3$ divided by $x$ square as $x$ tends to infinity, is obtained in the form of an arithmetic expression and it should be simplified to get the limit. So, let us concentrate on simplifying the arithmetic expression.

$=\,\,$ $2$ $+$ $3 \times 0^2$

$=\,\,$ $2$ $+$ $3 \times 0$

$=\,\,$ $2$ $+$ $0$

$=\,\,$ $2$

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