Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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
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
askIITians Faculty

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free