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

$x$ 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.

### Step: 1

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.

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

### Step: 2

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.

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

### Step: 3

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 $\Big( \dfrac{\pi}{x} \to 0 \Big)$.

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

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

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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