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I have tried to solve the following limit $$\lim\left(\frac{3y_n - x_n}{y^2_n}\right)$$ using the theorems for the limits, but I am not sure about what I am doing. Could you provide me some feedback and correct me (if I am wrong)?

We know that the limit of $x_n = 3$ and the limit of $y_n = 7$, and that all $y_n$ are non zero.

$$\lim\left(\frac{3y_n - x_n}{y^2_n}\right) = \frac{\lim(3y_n - x_n)}{\lim(y^2_n)} = \frac{3\lim(y_n) - \lim(x_n)}{\lim(y_n)\lim(y_n)} = \frac{3 \cdot 7 - 3}{49} = \frac{18}{49}$$

Is this correct? If not, where am I wrong? What can I improve?

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  • $\begingroup$ well since your limits don't do anything crazy, this is correct. The denominator can also be rewritten $(\lim y_n)^2$ because square is a continuous functions $\endgroup$ – Alex Apr 8 '15 at 11:31
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Hint:

If $(a_n)$ and $(b_n)$ convergent, therefore

$$\lim(a_n+b_n)=\lim a_n+\lim b_n$$ and $$\lim(a_n\cdot b_n)=\lim a_n\cdot \lim b_n.$$

It the case here, and thus, what you did is correct.

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  • $\begingroup$ In what cases of limits these techniques for finding the limits would not be useful and correct? $\endgroup$ – nbro Apr 8 '15 at 11:42
  • $\begingroup$ For exemple, ix you take $x_n=n$ and $y_n=-n$, then $$\lim (x_n+y_n)=0\neq \lim x_n +\lim y_n$$ because as you can see, $\lim x_n$ and $\lim y_n$ don't exist. $\endgroup$ – Surb Apr 8 '15 at 12:05
  • $\begingroup$ So, we must be careful about if each single limit exists or not, that is if the sequence converges to some number, because if not, we cannot apply these techniques.. $\endgroup$ – nbro Apr 8 '15 at 12:24
  • $\begingroup$ yes, you got it :-) $\endgroup$ – Surb Apr 8 '15 at 14:38

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