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Math  /  Calculus

Question8. Calculate the value of the following limit: lim⁡n→∞n4+n2−n4−5n2+n\lim _{n \rightarrow \infty} \sqrt{n^{4}+n^{2}}-\sqrt{n^{4}-5 n^{2}+n}

Studdy Solution

STEP 1

1. We are dealing with a limit as n n approaches infinity.
2. The expression involves square roots of polynomials.
3. Simplification techniques, such as rationalizing the expression, will be useful.

STEP 2

1. Simplify the expression inside the limit.
2. Rationalize the expression.
3. Evaluate the limit as n n approaches infinity.

STEP 3

First, observe the expression inside the limit:
lim⁡n→∞(n4+n2−n4−5n2+n) \lim_{n \to \infty} \left( \sqrt{n^4 + n^2} - \sqrt{n^4 - 5n^2 + n} \right)
To simplify, factor out n4 n^4 from each square root:
n4+n2=n4(1+1n2)=n21+1n2 \sqrt{n^4 + n^2} = \sqrt{n^4(1 + \frac{1}{n^2})} = n^2\sqrt{1 + \frac{1}{n^2}}
n4−5n2+n=n4(1−5n2+1n3)=n21−5n2+1n3 \sqrt{n^4 - 5n^2 + n} = \sqrt{n^4(1 - \frac{5}{n^2} + \frac{1}{n^3})} = n^2\sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}

STEP 4

Rationalize the expression by multiplying the numerator and the denominator by the conjugate:
lim⁡n→∞(n4+n2−n4−5n2+n)(n4+n2+n4−5n2+n)n4+n2+n4−5n2+n \lim_{n \to \infty} \frac{\left( \sqrt{n^4 + n^2} - \sqrt{n^4 - 5n^2 + n} \right) \left( \sqrt{n^4 + n^2} + \sqrt{n^4 - 5n^2 + n} \right)}{\sqrt{n^4 + n^2} + \sqrt{n^4 - 5n^2 + n}}
This simplifies to:
lim⁡n→∞(n4+n2)−(n4−5n2+n)n4+n2+n4−5n2+n \lim_{n \to \infty} \frac{(n^4 + n^2) - (n^4 - 5n^2 + n)}{\sqrt{n^4 + n^2} + \sqrt{n^4 - 5n^2 + n}}
=lim⁡n→∞n2+5n2−nn2(1+1n2+1−5n2+1n3) = \lim_{n \to \infty} \frac{n^2 + 5n^2 - n}{n^2\left(\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}\right)}
=lim⁡n→∞6n2−nn2(1+1n2+1−5n2+1n3) = \lim_{n \to \infty} \frac{6n^2 - n}{n^2\left(\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}\right)}

STEP 5

Divide each term by n2 n^2 in the numerator and simplify:
=lim⁡n→∞6−1n1+1n2+1−5n2+1n3 = \lim_{n \to \infty} \frac{6 - \frac{1}{n}}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}
As n→∞ n \to \infty , 1n→0 \frac{1}{n} \to 0 , 1n2→0 \frac{1}{n^2} \to 0 , and 1n3→0 \frac{1}{n^3} \to 0 . Therefore, the expression simplifies to:
=6−01+0+1−0+0 = \frac{6 - 0}{\sqrt{1 + 0} + \sqrt{1 - 0 + 0}}
=61+1 = \frac{6}{1 + 1}
=62 = \frac{6}{2}
=3 = 3
The value of the limit is:
3 \boxed{3}

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