Limit Continuity and Differentiability 3 Question 3

3. $\lim _{n \rightarrow \infty} \frac{1}{n} \sum _{r=1}^{2 n} \frac{r}{\sqrt{n^{2}+r^{2}}}$ equals

$(1999,2 M)$

(a) $1+\sqrt{5}$

(c) $-1+\sqrt{2}$

(b) $\sqrt{5}-1$

(d) $1+\sqrt{2}$

$(2007,3 M)$

Show Answer

Answer:

Correct Answer: 3. (a)

Solution:

  1. Given, $f(x)=\left[\tan ^{2} x\right]$

Now, $-45^{\circ}<x<45^{\circ}$

$$ \begin{array}{lc} \Rightarrow & \tan \left(-45^{\circ}\right)<\tan x<\tan \left(45^{\circ}\right) \\ \Rightarrow & -\tan 45^{\circ}<\tan x<\tan \left(45^{\circ}\right) \\ \Rightarrow & -1<\tan x<1 \\ \Rightarrow & 0<\tan ^{2} x<1 \\ \Rightarrow & {\left[\tan ^{2} x\right]=0} \end{array} $$

i.e. $f(x)$ is zero for all values of $x$ from $x=-45^{\circ}$ to $45^{\circ}$. Thus, $f(x)$ exists when $x \rightarrow 0$ and also it is continuous at $x=0$. Also, $f(x)$ is differentiable at $x=0$ and has a value of zero.

Therefore, (b) is the answer.



NCERT Chapter Video Solution

Dual Pane