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Limit of sum is sum of limits


Solution:

Part a

By Theorem 9.3, we have
$$
\lim(x_n+y_n)=\lim x_n+\lim y_n=3+7=10.
$$


Part b

By Theorems 9.2 and 9.3 we have
$$
\lim(3y_n-x_n)=3\lim y_n-\lim x_n=3 \cdot 7-3=18.
$$ By Theorem 9.4 we have
$$
\lim y_n^2=\left(\lim y_n\right)^2=7^2=49.
$$ It follows from Theorem 9.6 that
$$
\lim \frac{3y_n-x_n}{y_n^2}=\frac{\lim(3y_n-x_n)}{\lim y_n^2}=\frac{18}{49}.
$$


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