limits and functions

Q1)

When x-->0, then {x} also -->0

So we can take the limit as

\lim_{\left\{x \right\}\rightarrow 0}\frac{1}{\frac{tan\left\{x \right\}}{\left\{x \right\}}}

which turns out to be 1...

so i think (a) is right

@ani. as x--> 0_- {x} --> 1 isnt it? so limit shudnt exist naa?.. btw ans IS given 1 :(

asish second question is not clear.......can u rewrite it??

i think it was supposed to be
\lim_{n\rightarrow \infty}tan\alpha + \frac{tan \frac{\alpha }{2}}{2}+\frac{tan \frac{\alpha }{2^2}}{2^2}+.......+ \frac{tan \frac{\alpha }{2^n}}{2^n}

Q4

Observe that it is an even function so it will be symmetric about x=0 so

\frac{\partial y}{\partial x}=\frac{2x(1-ln(x^{2}+1)}{ln(x^{2}+1)^{2}}

At x=0 the function is not defined

i suspect that the min value will be when

(x^{2}+1)^{2}=e

we get

x=\sqrt{\sqrt{e}-1}