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# lim x---> 0  sec^-1(x/sinx)   does not exists ???  explain why

Jitender Singh IIT Delhi
6 years ago
Ans: Limit does exist
Sol:
$L = \lim_{x\rightarrow 0}sec^{-1}(\frac{x}{sinx})$
Lets first calculate left hand side limit =
$x\rightarrow -h; h\rightarrow 0$
$L_{1} = \lim_{h\rightarrow 0}sec^{-1}(\frac{-h}{sin(-h)})$
$L_{1} = \lim_{h\rightarrow 0}sec^{-1}(\frac{-h}{-sin(h)})$
$L_{1} = \lim_{h\rightarrow 0}sec^{-1}(\frac{h}{sin(h)})$
$sinh = h - \frac{h^{3}}{3!} + \frac{h^{5}}{5!} - .............$
$\lim_{h\rightarrow 0}(\frac{h}{sinh}) = 1$
$L_{1} = \lim_{h\rightarrow 0}sec^{-1}(1) = 0$
Lets calculate right side limit;
$x\rightarrow h; h\rightarrow 0$
$L_{2} = \lim_{h\rightarrow 0}(sec^{-1}(\frac{h}{sinh})) = \lim_{h\rightarrow 0}sec^{-1}(1) = 0$
$L_{1} = L_{2}$
Also, the value of the given function is also zero at x = 0
So, Limit exist at zero.
Cheers!
Thanks & Regards
Jitender Singh
IIT Delhi