Q33 / 50 · Manipal (MET) 2024 · Mathematics
Let $f(x)=\left[\frac{\sin x}{x}\right]+\left[\frac{2 \sin x}{x}\right]+\ldots+\left[\frac{10 \sin x}{x}\right]$ (where [.] is the largest integer). then, find the value of $\lim _{x \rightarrow 0} f(x)$.
Correct answer: 375
Solution
As we know,
$\lim _{x \rightarrow 0} \frac{\sin x}{x} \simeq 1$
but it is actually less than 1 .
So, $\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]=0=1^2-1$
Similarly, $\lim _{x \rightarrow 0}\left[2 \frac{\sin 2 x}{x}\right]=\lim _{x \rightarrow 0}\left[4 \frac{\sin 2 x}{2 x}\right]$
$=3=2^2-1$
Hence, $\lim _{x \rightarrow 0}\left[n \frac{\sin n x}{x}\right]=n^2-1$
Therefore,
$\begin{aligned}
& \lim _{x \rightarrow 0} f(x)=\left(1^2-1\right)+\left(2^2-1\right)+\ldots+\left(10^2-(1)\right. \\
& =\left(1^2+2^2+\ldots+10^2\right)-(1+1+\ldots+1 \text { times }) \\
& =\frac{10 \times 11 \times 21}{6}-10 \\
& =385-10=375
\end{aligned}$
Hence, the correct answer is 375.