Question

How to check AT SIGHT whether a limit exists or not? for ex.: lim(x->2){5/(rt(2)-rt(x))}

Answer 1

5
------------- if the top can cancel out the offending bottom
sqrt(2) - sqrt(x) then we can control the function

There is no way for '5' to control the bottom of this

Answer 2

there is a vertical asymptote at 2; not a hole.... if we had a whole, a limit would exist; if we had a hole we could rewrite it to an equavilant form without a hole

Answer 3

if both sides of the asymptote go to the same infinity tho, we might be justified in saying the limit is +- infinity

Answer 4

left hand limit goes to -infty and right hand limit is +infty.
therefore limit doesn't exist.
Now considering asymptotes is not "at sight" mathematics!!!

Answer 5

PS:u are not given the graph of this function

Answer 6

asymptotes is 'at sight' when you know what to look for in the equation...

Answer 7

a vertical asymptote occurs when you cant eliminate the offending zero by a like term in the numerator

Answer 8

x^2 -9
---------- has an issue qt -3; but that issue is resolved by the top
(x+3)(x+2)

Answer 9

(x+3)(x-3)
--------- ; the (x+3)s cancel out and we are left with just a hole
(x+3)(x+2)

Answer 10

if there is no way to control the offending zero in the bottom; we get an asymptote...

Answer 11

even if there is an asymptote how will you say both parts of the curve go to same infty i.e. both go to +infty/-infty or one goes to + and other to -infty?

Answer 12

a sign chart helps :) You know what the zeros are; so plot out there signs on a chart and you will see which side is neg and pos

Answer 13

if there is only one zero; its most likely opposites, but that aint a general rule because its determined by the equation

Answer 14

<..... 2 ......>
- 0 +
------------
-inf VA inf ; for this one tho