Question

Solve the following. $\mathop {\lim }\limits_{n \to \infty } \sin \left[ {\pi \sqrt {{n^2} + 1} } \right]$ is equal to
A) $\infty$
B) 0
C) Does not exist
D) None of above

Answer 1

In this question, we will apply the trigonometry formula $\sin x = {\left( { - 1} \right)^{n + 1}}\sin \left[ {n\pi - x} \right]$. Because when we apply the limit over the given function, it will give $\sin \infty$. So, we have to check for two cases of$\sin (n\pi - x)$. The first case is n is even. When we put the even values of n they will lie in the fourth quadrant. In the fourth quadrant, the value of the sine function is negative. Therefore, $\sin (n\pi - x) = - \sin x$. The second case is odd. When we put the odd values of n they will lie in the second quadrant. In the second quadrant, the value of sin function is positive. Therefore, $\sin (n\pi - x) = \sin x$. Hence, we can write $\sin (n\pi - x) = {\left( { - 1} \right)^{n + 1}}\sin x$ in a generalized form.
Then apply rationalization to the numerator. Here, we will apply rationalization because when we will apply the limit to the function we will get the value $\sin \left( {\infty - \infty } \right)$, which is not possible.
The other algebraic formula that we will apply here during simplification is as stated below.
$\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$

Complete step by step solution:
In this question, it is given that
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \sin \left[ {\pi \sqrt {{n^2} + 1} } \right]$
As we know the trigonometry formula $\sin x = {\left( { - 1} \right)^{n + 1}}\sin \left[ {n\pi - x} \right]$.
Here, substitute the value of x is equal to $\pi \sqrt {{n^2} + 1}$.
Therefore, we will get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \left( {n\pi - \pi \sqrt {{n^2} + 1} } \right)$
Take out $\pi$ as a common factor. Because it has a constant value.
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {n - \sqrt {{n^2} + 1} } \right)$
When we substitute value $n = \infty$, we will get $\sin \left( {\infty - \infty } \right)$. That is not possible.
Therefore, Let us rationalize the numerator of the above step.
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{\left( {n - \sqrt {{n^2} + 1} } \right)\left( {n + \sqrt {{n^2} + 1} } \right)}}{{n + \sqrt {{n^2} + 1} }}} \right)$
Let us apply the formula $\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$.
Here, $a = n$ and $b = \sqrt {{n^2} + 1}$
So,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{{n^2} - {{\left( {\sqrt {{n^2} + 1} } \right)}^2}}}{{n + \sqrt {{n^2} + 1} }}} \right)$
Let us simplify the above step.
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{{n^2} - ({n^2} + 1)}}{{n + \sqrt {{n^2} + 1} }}} \right)$
Now, open the brackets and multiply with -1 with bracket values.
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{{n^2} - {n^2} - 1}}{{n + \sqrt {{n^2} + 1} }}} \right)$
By applying subtraction,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{ - 1}}{{n + \sqrt {{n^2} + 1} }}} \right)$
When we substitute value $n = \infty$, then we will get,
$\Rightarrow {\left( { - 1} \right)^{\infty + 1}}\sin \pi \left( {\dfrac{{ - 1}}{\infty }} \right)$
As we know that the value of $\dfrac{1}{\infty }$ is 0.
Therefore,
$\Rightarrow {\left( { - 1} \right)^{\infty + 1}}\sin \pi \left( 0 \right)$
0 multiply with any number the answer will be 0.
$\Rightarrow {\left( { - 1} \right)^{\infty + 1}}\sin 0$
We know the value of $\sin 0$ is 0.
Therefore, the answer will be,
$\Rightarrow 0$
Hence,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \sin \left[ {\pi \sqrt {{n^2} + 1} } \right] = 0$

Option B is the correct answer.

Note: We have to remember the trigonometry formula and trigonometry ratio values at the angle $0^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ$. And we have to remember the trigonometric function values in all quadrants.
In the first quadrant, sine and cosine functions are positive.
In the second quadrant, the sine function is positive and the cosine function is negative.
In the third quadrant, sine and cosine functions are negative.
In the fourth quadrant, the cosine function is positive and the sine function is negative.