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Math  /  Calculus

QuestionCalculate the value of the following limit: lim⁡n→∞3n2+n−3n2−5n\lim _{n \rightarrow \infty} \sqrt{3 n^{2}+n}-\sqrt{3 n^{2}-5 n}

Studdy Solution

STEP 1

1. The limit involves expressions with square roots and polynomial terms.
2. As n n approaches infinity, the dominant terms in the square roots will determine the behavior of the expression.
3. Rationalizing the expression may help simplify the limit calculation.

STEP 2

1. Simplify the expression by rationalizing the difference of square roots.
2. Evaluate the limit of the simplified expression as n n approaches infinity.

STEP 3

To simplify the expression, multiply and divide by the conjugate of the square roots:
lim⁡n→∞(3n2+n−3n2−5n)\lim _{n \rightarrow \infty} \left( \sqrt{3 n^{2}+n} - \sqrt{3 n^{2}-5 n} \right)
Multiply and divide by the conjugate:
lim⁡n→∞(3n2+n−3n2−5n)(3n2+n+3n2−5n)3n2+n+3n2−5n\lim _{n \rightarrow \infty} \frac{\left( \sqrt{3 n^{2}+n} - \sqrt{3 n^{2}-5 n} \right) \left( \sqrt{3 n^{2}+n} + \sqrt{3 n^{2}-5 n} \right)}{\sqrt{3 n^{2}+n} + \sqrt{3 n^{2}-5 n}}

STEP 4

Simplify the numerator using the difference of squares formula:
(3n2+n)2−(3n2−5n)2=(3n2+n)−(3n2−5n)\left( \sqrt{3 n^{2}+n} \right)^2 - \left( \sqrt{3 n^{2}-5 n} \right)^2 = (3 n^{2} + n) - (3 n^{2} - 5 n)
Simplify further:
=3n2+n−3n2+5n=6n= 3n^{2} + n - 3n^{2} + 5n = 6n
The expression becomes:
lim⁡n→∞6n3n2+n+3n2−5n\lim _{n \rightarrow \infty} \frac{6n}{\sqrt{3 n^{2}+n} + \sqrt{3 n^{2}-5 n}}

STEP 5

Factor out n n from the square roots in the denominator:
3n2+n=n3+1n\sqrt{3 n^{2}+n} = n\sqrt{3 + \frac{1}{n}} 3n2−5n=n3−5n\sqrt{3 n^{2}-5 n} = n\sqrt{3 - \frac{5}{n}}
The expression becomes:
lim⁡n→∞6nn(3+1n+3−5n)\lim _{n \rightarrow \infty} \frac{6n}{n\left(\sqrt{3 + \frac{1}{n}} + \sqrt{3 - \frac{5}{n}}\right)}
Cancel n n in the numerator and denominator:
lim⁡n→∞63+1n+3−5n\lim _{n \rightarrow \infty} \frac{6}{\sqrt{3 + \frac{1}{n}} + \sqrt{3 - \frac{5}{n}}}

STEP 6

Evaluate the limit as n→∞ n \rightarrow \infty :
As n→∞ n \rightarrow \infty , 1n→0 \frac{1}{n} \rightarrow 0 and 5n→0 \frac{5}{n} \rightarrow 0 , so:
3+1n→3\sqrt{3 + \frac{1}{n}} \rightarrow \sqrt{3} 3−5n→3\sqrt{3 - \frac{5}{n}} \rightarrow \sqrt{3}
Thus, the expression simplifies to:
lim⁡n→∞63+3=623=33=3\lim _{n \rightarrow \infty} \frac{6}{\sqrt{3} + \sqrt{3}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}
The value of the limit is:
3 \boxed{\sqrt{3}}

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