$x$ and $\sin \Big(\dfrac{\pi}{x}\Big)$ are two functions and the value of product of them is required to find when $x$ approaches infinity.

$$\lim_{x \to \infty} x \sin \Big( \dfrac{\pi}{x} \Big)$$

Observe the formation of the function. The angle of the sine function is $\dfrac{\pi}{x}$ and $x$ is outside of the function. If the variable $x$ is used to transform it as angle of the sine function. The ratio of sine of angle to angle limit identity can be used to solve this problem.

In this limits problem, the two functions $x$ and $\sin\Big(\dfrac{\pi}{x}\Big)$ are multiplying each other to form another function by their product.

$$= \lim_{x \to \infty} x \times \sin \Big( \dfrac{\pi}{x} \Big)$$

Try to adjust the function to create a denominator which should be equal to the angle of the sine function and follow below steps to do it successfully.

Mathematically, the literal $x$ can be expressed as the product of $x$ and $1$.

$$= \lim_{x \to \infty} x \times 1 \times \sin \Big( \dfrac{\pi}{x} \Big)$$

Write the number $1$ as the quotient of the ratio of $\pi$ to $\pi$.

$$= \lim_{x \to \infty} x \times \dfrac{\pi}{\pi} \times \sin \Big( \dfrac{\pi}{x} \Big)$$

Change the numerator of the fraction by the multiplication rule of fractions.

$$= \lim_{x \to \infty} \pi \times \dfrac{x}{\pi} \times \sin \Big( \dfrac{\pi}{x} \Big)$$

The fraction $\dfrac{x}{\pi}$ is multiplying the remaining factors and its reciprocal divides the same multiplying factors. Hence, move the fraction to denominator and make $\sin \Big(\dfrac{\pi}{x}\Big)$ as numerator for that fraction.

$$= \lim_{x \to \infty} \pi \times \dfrac{\sin \Big( \dfrac{\pi}{x} \Big)}{\dfrac{\pi}{x}}$$

The $\pi$ is a constant and take out from the limit expression.

$$= \pi \times \lim_{x \to \infty} \dfrac{\sin \Big( \dfrac{\pi}{x} \Big)}{\dfrac{\pi}{x}}$$

When $x$ tends to infinity ($x \to \infty$), then the ratio of $1$ to $x$ approaches zero $\Big( \dfrac{1}{x} \to 0 \Big)$. Similarly, the value of ratio of $\pi$ to $x$ also tends to zero $\Big( \dfrac{\pi}{x} \to 0 \Big)$.

$$= \pi \times \lim_{\displaystyle \dfrac{1}{x} \to 0} \dfrac{\sin \Big( \dfrac{\pi}{x} \Big)}{\dfrac{\pi}{x}}$$

$$= \pi \times \lim_{\displaystyle \dfrac{\pi}{x} \to 0} \dfrac{\sin \Big( \dfrac{\pi}{x} \Big)}{\dfrac{\pi}{x}}$$

Now, apply limit angle tends to zero, the value of ratio of sine of angle to angle is one rule to solve this problem.

$= \pi \times 1$

$$\therefore \,\,\,\,\,\, \lim_{x \to \infty} x \sin \Big( \dfrac{\pi}{x} \Big) = \pi$$

Therefore, it is proved that the value of product of $x$ and $\sin \Big( \dfrac{\pi}{x} \Big)$ when $x$ approaches infinity is $\pi$.

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