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

Last updated date: 11th Aug 2024
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Hint: 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.
$\Rightarrow 0$
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \sin \left[ {\pi \sqrt {{n^2} + 1} } \right] = 0$
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.