Question

How do you determine the limit of \[\dfrac{{\sin (x)}}{x}\] as x approaches infinity?

Answer 1

Hint: According to given in the question we have to determine the limit of the given trigonometric function which is as given in the question is \[\dfrac{{\sin (x)}}{x}\] as x approaches infinity. So, first of all to find the limit of the given trigonometric function as x approaches infinity, the y-values oscillates between 1 and -1 and as we know that this limit does not exist.
Now, to determine the given limit we have to solve that limit is that generally, if there is no x in the denominator of the given function at all, then we can say that the limit does not exist.
Now, to solve the limit we have to use the squeeze theorem which is as explained below:
Squeeze theorem: It is also known as the pinching theorem or the sandwich theorem which is a theorem regarding the limit of a function. It is used to confirm the limit of a function via comparison between two other functions whose limits are known or we can say that whose limits can be easily computed.
Now, we have to recall that $\sin (x)$ is only defined on the $ - 1 \leqslant \sin x \leqslant 1$ therefore, we have to find the limit for the given function \[\dfrac{{\sin (x)}}{x}\].
Now, we to determine the limits for ${\lim _{x \to \infty }} - \dfrac{1}{x}$ and ${\lim _{x \to \infty }}\dfrac{1}{x}$ to determine the required limit of the given function.

Complete step-by-step solution:
Step 1: First of all to find the limit of the given trigonometric function as x approaches infinity, the y-values oscillates between 1 and -1 and as we know that this limit does not exist which is as explained in the solution hint.
Step 2: Now, to determine the given limit we have to solve that limit is that generally, if there is no x in the denominator of the given function at all, then we can say that the limit does not exist.
Step 3: Now, we have to use the squeeze theorem and recall that $\sin (x)$is only defined on the $ - 1 \leqslant \sin x \leqslant 1$therefore, we have to find the limit for the given function \[\dfrac{{\sin (x)}}{x}\]. Which is already explained in the solution hint.
$ \Rightarrow - \dfrac{1}{x} \leqslant \dfrac{{\sin (x)}}{x} \leqslant \dfrac{1}{x}$
Step 4: Now, we to determine the limits for ${\lim _{x \to \infty }} - \dfrac{1}{x}$and ${\lim _{x \to \infty }}\dfrac{1}{x}$to determine the required limit of the given function. Hence,
$
   = {\lim _{x \to \infty }} - \dfrac{1}{x} \\
   = {\lim _{x \to \infty }} - \dfrac{1}{0} \\
   = 0
 $
And,
$
   = {\lim _{x \to \infty }}\dfrac{1}{x} \\
   = {\lim _{x \to \infty }}\dfrac{1}{0} \\
   = 0
 $
Step 5: Hence, with the help of step 4 we can determine the value of limit for the given fi=unction which is as below:
$ \Rightarrow {\lim _{x \to \infty }}\dfrac{{\sin (x)}}{x} = 0$

Hence, with the help of the squeeze theorem we have determined the limit of the given function \[\dfrac{{\sin (x)}}{x}\] as x approaches infinity is 0.

Note: The squeeze lemma or the squeeze theorem is used in calculus to evaluate the limit of a function and this theorem is practically used to evaluate the limits of the given function where the other methods are unnecessarily difficult.
It is used to confirm the limit of a function via comparison between two other functions whose limits are known or we can say that whose limits can be easily computed.