lim x ->(+infinity) [sqrt(x² + 8x + 50) - sqrt(x + 2)]?

Question

Help pls.!
Tulong naman diyan!

^_^

Answer 1

Here ya go buddy. As x tends to infinity, only the highest power terms will dominate. All you have to do is compare the highest power terms in under each root. Effectively, the highest power of x in the first term is x^1 (since sqrt(x^2)=x) and the highest power in the second term is sqrt(x).

Now, as x->inf, you are effectively looking at the function x-sqrt(x). Since there are no denominator terms, this function goes straight to infinity.

Also, you could multiply by the conjugate to get
(x^2+7x+50)/((sqrt(x^2+8x+50)+sqrt(x+2))
but you'd still have to go through the same kind of rationale, where you would be dealing with:
lim x->inf x^2/x = x
which still goes to infinity. Only use the conjugate if you can't make this kind of connection or if you are evaluating your limit at a finite value.

Answer 2

You just need to appreciate what happens to each set of expressions within this one - what happens when X gets very large. Which number will be larger? sqrt(x² + 8x + 50)? or sqrt(x + 2?
Once this has been established you can say what number/expression this tends to when x tends to infinity.

Answer 3

If x is better than 4, sqrtx would be decrease than x/2, and hence x-sqrtx would be better than x/2. Now x/2 could be made arbitrarily super as you boost x, so your expression additionally gets arbitrarily super as x is going to infinity. it fairly is to assert "the cut back is infinity".

Answer 4

try taking the cojucate of the term this way the above sqrt will be removed and then you can solve it

Answer 5

There is a simple trick in such cases. You use a-b= (a2-b2)/(a+b)
and then it is simpler.