I'm trying to take limit as x approach 0 of ((sqrt(x+7)+sqrt(7))/x)
I thought it would be infinity but when I put it in wolframalpha it takes limit from right / left so the answer is negative infinity and positive infinity.
How do you know when to look at limit from right and left?
I thought it would be infinity but when I put it in wolframalpha it takes limit from right / left so the answer is negative infinity and positive infinity.
How do you know when to look at limit from right and left?
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When you are finding lim(x->a){f(x)} and f is discontinuous at a, then you must consider the left-hand and right-hand limits separately.
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the fact that the limits are different as x -> 0 from either end means that
, in theory, the limit does NOT exist.
to test use x = 0 + h or x = 0 -h where h >0
get lt h -> 0 : [√(h+7)+√(7)]/h = [√7√(1+h/7)+√(7)]/h = √(7). [√(1+h/7) +1] /h
~ √(7). [(1+h/14) +1] /h = √(7). (2+h/14) /h ~ 2√(7)/h i.e. + inf.
now let x = 0 -h wher h>0 also
√(7). (2-h/14) /(-h) ~ -2√(7)/h i.e. - inf.
both limits diverge one +ve and one -ve meaning that at x =0 the limit can't exist!
If we were talking about [√(x+7)-√(7)]/x by the way the limit would equate and so exist
as √(7)/14 in both cases - are you sure you have wrote the problem down correctly?
, in theory, the limit does NOT exist.
to test use x = 0 + h or x = 0 -h where h >0
get lt h -> 0 : [√(h+7)+√(7)]/h = [√7√(1+h/7)+√(7)]/h = √(7). [√(1+h/7) +1] /h
~ √(7). [(1+h/14) +1] /h = √(7). (2+h/14) /h ~ 2√(7)/h i.e. + inf.
now let x = 0 -h wher h>0 also
√(7). (2-h/14) /(-h) ~ -2√(7)/h i.e. - inf.
both limits diverge one +ve and one -ve meaning that at x =0 the limit can't exist!
If we were talking about [√(x+7)-√(7)]/x by the way the limit would equate and so exist
as √(7)/14 in both cases - are you sure you have wrote the problem down correctly?