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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}
ANS:

Studdy Solution

STEP 1

1. The expression involves two square roots and a limit as n n approaches infinity.
2. Simplification of the expression is necessary to evaluate the limit.
3. Rationalization techniques may be useful in simplifying the expression.

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 \rightarrow \infty} \left( \sqrt{n^{4}+n^{2}} - \sqrt{n^{4}-5n^{2}+n} \right)
Notice that both terms under the square roots have n4 n^4 as the leading term. We can factor n4 n^4 out of each square root:
n4(1+1n2)−n4(1−5n2+1n3) \sqrt{n^4(1 + \frac{1}{n^2})} - \sqrt{n^4(1 - \frac{5}{n^2} + \frac{1}{n^3})}
This simplifies to:
n21+1n2−n21−5n2+1n3 n^2 \sqrt{1 + \frac{1}{n^2}} - n^2 \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}

STEP 4

To simplify further, we rationalize the expression by multiplying and dividing by the conjugate:
(n21+1n2−n21−5n2+1n3)×1+1n2+1−5n2+1n31+1n2+1−5n2+1n3 \left( n^2 \sqrt{1 + \frac{1}{n^2}} - n^2 \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}} \right) \times \frac{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}
This results in:
n2((1+1n2)−(1−5n2+1n3))1+1n2+1−5n2+1n3 \frac{n^2 \left( \left(1 + \frac{1}{n^2}\right) - \left(1 - \frac{5}{n^2} + \frac{1}{n^3}\right) \right)}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}
Simplifying the numerator:
n2(1n2+5n2−1n3)=n2(6n2−1n3)=6−1n n^2 \left( \frac{1}{n^2} + \frac{5}{n^2} - \frac{1}{n^3} \right) = n^2 \left( \frac{6}{n^2} - \frac{1}{n^3} \right) = 6 - \frac{1}{n}

STEP 5

Now, evaluate the limit as n n approaches infinity:
lim⁡n→∞6−1n1+1n2+1−5n2+1n3 \lim _{n \rightarrow \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 , the terms 1n2 \frac{1}{n^2} , 5n2 \frac{5}{n^2} , and 1n3 \frac{1}{n^3} approach zero, so:
1+1n2→1 \sqrt{1 + \frac{1}{n^2}} \to 1 1−5n2+1n3→1 \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}} \to 1
Thus, the denominator approaches 1+1=2 1 + 1 = 2 .
The limit becomes:
lim⁡n→∞6−1n2=62=3 \lim _{n \rightarrow \infty} \frac{6 - \frac{1}{n}}{2} = \frac{6}{2} = 3
The value of the limit is:
3 \boxed{3}

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