0

Guest

Courses New

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

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

Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 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
askIITians Faculty

Think You Can Provide A Better Answer ?

Other Related Questions on Discuss with Askiitians Tutors

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More