Find the limit of the sequence if it converges; otherwise indicate divergence.
-$$a _ { n } = n - \sqrt { n ^ { 2 } - 15 n }$$
A) 0
B) $\sqrt { 15 }$
C) $\frac { 15 } { 2 }$
D) Diverges

Question

Quizplus · Accepted Answer

We can simplify $a_n$ as follows:
$$a_n = n - \sqrt{n^2-15n} = n - \sqrt{n(n-15)} = \frac{n(\sqrt{n(n-15)}-n+15)}{\sqrt{n(n-15)}+n-15}$$
Now, we can see that both the numerator and denominator tend to infinity as $n$ increases. Therefore, by applying L'Hopital's rule, we get:
$$\lim_{n	o\infty}a_n = \lim_{n	o\infty}\frac{\sqrt{n(n-15)}+15}{\sqrt{n(n-15)}+n-15} = \lim_{n	o\infty}\frac{\frac{1}{2\sqrt{n(n-15)}}(2n-15)}{\frac{1}{2\sqrt{n(n-15)}}(2n-15)+1} = \frac{\frac{1}{2}}{\frac{1}{2}+1} = \frac{1}{3}\cdot 15 = \frac{15}{3} = \boxed{\frac{15}{2}}$$

Find the Limit of the Sequence If It Converges; Otherwise an=n−n2−15na _ { n } = n - \sqrt { n ^ { 2 } - 15 n }an​=n−n2−15n​

Find the Limit of the Sequence If It Converges; Otherwise $a _ { n } = n - \sqrt { n ^ { 2 } - 15 n }$