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Determine the convergence or divergence of the sequence an=nSin(1/n), and if its convergent, find its limit?

Answer 1

Emily Ammerman

Converges to #1.#

If a sequence #a_n# converges, #lim_(n->oo)a_n=L# where #L# is some constant (non-infinite) value.

If #lim_(n->oo)a_n=+-oo,# the sequence diverges. So, let's take the limit:

#lim_(n->oo)nsin(1/n)=oosin(1/oo)=oosin(0)=oo(0)#

This is an ambiguous form that provides little information.

Now, we can rewrite as #a_n=sin(1/n)/(1/n)# and plan on using l'Hospital's Rule; however, a sequence is not differentiable, so, we must define a new function

#f(x)=sin(1/x)/(1/x)# which we can differentiate to use l'Hospital's Rule on.

#lim_(x->oo)sin(1/x)/(1/x)=((cancel(-1/x^2))cos(1/x))/(cancel(-1/x^2))=lim_(x->oo)cos(1/x)=cos(0)=1#

Then, the sequence converges to #1#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.