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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..
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).