I'm stuck trying to find the limit of the sequence √12+an−√4ana2n−2an−8
Where I'm given that an>4 and an→4
Both the numerator and the denominator tend to 0, and I can't find how to solve this indetermination. I tried multiplying and dividing by the "reciprocal" of the numerator to get rid of the square roots in the numerator, but that doesn't seem to lead anywhere. What else can I try?
Answer
Hint:
b2−2b−8=(b−4)(b+2)
√12+b−√4b=−3(b−4)√12+b+√4b
If b→4,b≠4⟹b−4≠0 hence can be cancelled safely
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