Studdy AI Logo
Math  /  Calculus

Question8. Calculate the value of the following limit:
ANS: \qquad 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 limits as n n approaches infinity.
2. The dominant term in both square roots is n4 n^4 .
3. Simplification will involve rationalizing the expression.

STEP 2

1. Simplify the expression inside the limit.
2. Rationalize the expression.
3. Evaluate the limit as n→∞ n \to \infty .

STEP 3

First, consider the expression:
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:
=lim⁡n→∞(n4(1+1n2)−n4(1−5n2+1n3)) = \lim_{n \to \infty} \left( \sqrt{n^4(1 + \frac{1}{n^2})} - \sqrt{n^4(1 - \frac{5}{n^2} + \frac{1}{n^3})} \right)
=lim⁡n→∞(n21+1n2−n21−5n2+1n3) = \lim_{n \to \infty} \left( n^2\sqrt{1 + \frac{1}{n^2}} - n^2\sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}} \right)

STEP 4

To simplify further, factor n2 n^2 out of the expression:
=n2lim⁡n→∞(1+1n2−1−5n2+1n3) = n^2 \lim_{n \to \infty} \left( \sqrt{1 + \frac{1}{n^2}} - \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}} \right)
Rationalize the expression by multiplying and dividing by the conjugate:
=n2lim⁡n→∞(1+1n2−1−5n2+1n3)(1+1n2+1−5n2+1n3)1+1n2+1−5n2+1n3 = n^2 \lim_{n \to \infty} \frac{\left( \sqrt{1 + \frac{1}{n^2}} - \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}} \right) \left( \sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}} \right)}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}

STEP 5

Simplify the numerator using the difference of squares:
=n2lim⁡n→∞(1+1n2)−(1−5n2+1n3)1+1n2+1−5n2+1n3 = n^2 \lim_{n \to \infty} \frac{\left( 1 + \frac{1}{n^2} \right) - \left( 1 - \frac{5}{n^2} + \frac{1}{n^3} \right)}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}
=n2lim⁡n→∞1n2+5n2−1n31+1n2+1−5n2+1n3 = n^2 \lim_{n \to \infty} \frac{\frac{1}{n^2} + \frac{5}{n^2} - \frac{1}{n^3}}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}
=n2lim⁡n→∞6n2−1n31+1n2+1−5n2+1n3 = n^2 \lim_{n \to \infty} \frac{\frac{6}{n^2} - \frac{1}{n^3}}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}

STEP 6

Simplify the expression further and evaluate the limit:
=n2⋅6n2⋅lim⁡n→∞1−16n1+1n2+1−5n2+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:
=6⋅12=3 = 6 \cdot \frac{1}{2} = 3
The value of the limit is:
3 \boxed{3}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord