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depends on the problem...
use limits for that, say lim as x goes to infinity of x-x = 0, so u can't be sure
i believe it is set as an inequality, for example, 4 + \infty = 4+ \infty i > \infty i
also, you can use limits and logarithms to solve the solution
*i believe it is set as an inequality, for example, 4 + infty = 4+ infty > infty
hey mathfreak, im busy right now, help out the problem below this.
can i please get help with this problem 2 2/3 + 3 5/6 + 9 1/4, it's easy... i just gotta do my own hw
nevermind, someone's helpin, i gtg now ttyl folks
i am so confuse. for example, lim x-> 3+ (1/x-3)+10 the answer is = infinitive. why its not 10...
i will take a look later tonite.. very busy now, sorry
oh i know why its infinity, will explain later
The answer is infinity because when X becomes 3 the bottom of the fraction equals zero. When you divide 1 by a really small number the value is really big because a really small value goes into something many times. Then as 1/0 is undefined so is infinity. So when x goes to three it is going to infinity.
Let me elaborate on this problem: You have lim as x -> 3+ of ( 1/(x-3) -10 ) = infinity. Why? The answer above is basically correct, because 1/(x-3) approaches infinity, as x approaches 3 from the right side. Pick some number that approaches 3 from the right, say 3.00000001: 1/(3.00000001-3) = 1,000,000,000. The smaller the number becomes, the larger the limit: For 3.000000000001 it would be 10,000,000,000,000; etc, etc..
so in that case it approaches infinity
What if x approached three from the left, or you had lim as x -> 3- of ( 1/(x-3) -10 ). Pick some number that approaches 3 from the left, say 2.9999999, then your answer would approach negative infinity, as you would be getting -100000000000000, etc etc instead of +10000000000000 etc, etc. And in both cases you can ignore the -10, as infinity(some very large number) minus 10 is still infinity(some very large number), so –infinity – 10 is still –infinity, because 10 is insignificant once we are dealing with huge numbers.
Now if you had to solve lim as x -> 3 of ( 1/(x-3) -10 ), just three, without any plus or minus above it, the answer would be the limit does not exist, as I just showed that if I approached 3 from the right, I would get +infinity and if I approached 3 from the left I would get –infinity. For limit to exist, right hand limit must be equal to left hand limit, so in case of lim as x-> of ( 1/(x-3) -10) the limit would NOT exist; it would not be equal to neither –infinity or +infinity (it would be equal to +infinity if limit from left = limit from right = +infinity, or –infinity if limit from left = limit from right = -infinity).
limits are kinda hard to understand, but it's more of abstract type of stuff... but i think u will get it eventually.