MHB More Convergence & Divergence with sequences

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The sequence \( a_n = n \sin(1/n) \) converges to 1 as \( n \) approaches infinity. While using l'Hôpital's rule is one approach, it is more effective to apply the fundamental limit \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \) by substituting \( x = \frac{1}{n} \). This method avoids unnecessary complications and provides a clearer path to the limit. The discussion emphasizes the importance of recognizing fundamental limits in calculus. Ultimately, the sequence converges to 1.
shamieh
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Determine whether the sequence converges or diverges, if it converges fidn the limit.

$$a_n = n \sin(1/n)$$

so Can I just do this:

$$n * \sin(1/n)$$ is indeterminate form

so i can use lopitals

so:

$$1 * \cos(1/x) = 1 * 1 = 1$$

converges to 1?
 
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shamieh said:
Determine whether the sequence converges or diverges, if it converges fidn the limit.

$$a_n = n \sin(1/n)$$

so Can I just do this:

$$n * \sin(1/n)$$ is indeterminate form

so i can use lopitals

so:

$$1 * \cos(1/x) = 1 * 1 = 1$$

converges to 1?

Your result is correct... however, if You want to use continuous functions, it is better to set $\displaystyle x=\frac{1}{n}$ and the limit becomes $\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}=1$. This is a 'fundamental limit' and the l'Hopital rule shouldn't be used...

Kind regards

$\chi$ $\sigma$
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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