limn→∞√n+1+3√n+2−4
I know that I should multiply by the conjugate if I have square roots in either the numerator or the denominator but what if I have square roots in both numerator and denominator?
Could anyone help me please?
Answer
HINT
By intuition we have that the √n terms are dominant on the others and therefore in the limit
√(n+1)+3√(n+2)−4∼√n√n=1
To check and formalize rigoursly this fact let consider
√(n+1)+3√(n+2)−4=√n√n√(1+1/n)+3/√n√(1+2/n)−4/√n
then take the limit.
No comments:
Post a Comment