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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
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
if both sides of the asymptote go to the same infinity tho, we might be justified in saying the limit is +- infinity
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!!!
PS:u are not given the graph of this function
asymptotes is 'at sight' when you know what to look for in the equation...
a vertical asymptote occurs when you cant eliminate the offending zero by a like term in the numerator
x^2 -9 ---------- has an issue qt -3; but that issue is resolved by the top (x+3)(x+2)
(x+3)(x-3) --------- ; the (x+3)s cancel out and we are left with just a hole (x+3)(x+2)
if there is no way to control the offending zero in the bottom; we get an asymptote...
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?
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
if there is only one zero; its most likely opposites, but that aint a general rule because its determined by the equation
<..... 2 ......> - 0 + ------------ -inf VA inf ; for this one tho