Could someone help me through this problem?
Calculate limn→+∞(√n2+n−n)
Answer
We have:
√n2+n−n=(√n2+n−n)(√n2+n+n)√n2+n+n=n√n2+n+n
Therefore:
√n2+n−n=1√1+1n+1
And since: limn→+∞1n=0
It follows that:
limn→+∞(√n2+n−n)=12
I have injection f:A→B and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...
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