LIMITS Again

well write sinx as x

ull get it as (1-xcosx)/x²

applying the limits it is 1/0₊ = +∞

but r u allowed to write sinx =x

i guess ans is 0....

(1-x²cotx)/x²

now use exp of cot x
and then apply l hospital

its will be ∞.
Write cot x = 1/(tan x), take LCM & then apply L'Hospital's Rule.....if u want the full solution to convince urself.....well it looks just by seeing the q that it is ∞ !

how do u exp cot x?

yes virang when x--> 0 sinx/x --> 1 so sinx --> x

so ur allowed to rite that... (but in some cases where u have sinx - tanx) dont use it... tho..

ok u can directly use l hopital der

or may use this xcotx=1 -x³/3 + x⁴/45...............

and then apply l hospital

Limit does not exist.

how come does not exist ??

LHL = (∞ + ∞), which is always +ve.
But, RHL = (∞ - ∞), which can be +ve or -ve!
Hence, the limit cant exist!! :)