Math  /  Calculus

Question8. Calculate the value of the following limit:
ANS: \qquad limnn4+n2n45n2+n\lim _{n \rightarrow \infty} \sqrt{n^{4}+n^{2}}-\sqrt{n^{4}-5 n^{2}+n}
ANS:

Studdy Solution
Simplify the expression further and evaluate the limit:
=n26n2limn116n1+1n2+15n2+1n3 = n^2 \cdot \frac{6}{n^2} \cdot \lim_{n \to \infty} \frac{1 - \frac{1}{6n}}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}
As n n \to \infty , the terms 1n2\frac{1}{n^2}, 5n2\frac{5}{n^2}, and 1n3\frac{1}{n^3} approach zero:
=612=3 = 6 \cdot \frac{1}{2} = 3
The value of the limit is:
3 \boxed{3}

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