Question

What is $\displaystyle \lim_{x \to \infty }\dfrac{\sin 2x}{x}$ ?

Answer 1

This problem requires intuition. We have to note here that despite x tending to infinity, the value of $\sin 2x$ remains within $-1$ and $1$ . So, the numerator remains finite but the denominator tends to infinity. This gives the limit as $0$ .

Complete step-by-step answer:
In mathematics, limit is the value or amount that a function or sequence or series approaches or tends to as the variable or input approaches or tends to some other value. In limits, the variable or the function does not attain the destined value, but shows a behaviour to attain it. For example, the graph of $xy=c$ is not defined at $x=0$ . But we can get a clear idea about the behaviour of the graph. As the value of x approaches zero, the value of the function increases continuously. So, we can say that the function approaches infinity as x tends to zero.
Now, $\sin x$ or $\sin 2x$ are periodic functions with their domain being $\left( -\infty ,\infty \right)$ but their range being limited between $-1$ and $1$ . This means that whatever be the value of the variable x, the function always stays within its boundaries, the boundaries being $-1$ and $1$ , no matter how large the value of x. But the denominator of the limit function tends to infinity or a very large value. The function tends to the form $\dfrac{c\in \left[ -1,1 \right]}{\infty }$ which is nothing but $0$ .
Thus, we can conclude that $\displaystyle \lim_{x \to \infty }\dfrac{\sin 2x}{x}=0$ .

Note: The most common and grave mistake that students make here is that they apply the L’Hospital’s rule, even though the function does not tend to the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ . Also, some of us try to express $\sin 2x$ as an infinite series using Maclaurin series expansion. Even that process does not yield any results.