Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To find the limit of the function [tex]\(\lim_{{n \to 2}} \frac{4-n^2}{3-\sqrt{n^2+5}}\)[/tex], we need to follow these steps:
1. Substitute the value where [tex]\( n \)[/tex] approaches into the expression:
Substitute [tex]\( n = 2 \)[/tex] into the function:
[tex]\[ \frac{4 - (2)^2}{3 - \sqrt{(2)^2 + 5}} \][/tex]
Evaluating inside the parentheses:
[tex]\[ \frac{4 - 4}{3 - \sqrt{4 + 5}} = \frac{0}{3 - \sqrt{9}} \][/tex]
Since [tex]\(\sqrt{9} = 3\)[/tex], we get:
[tex]\[ \frac{0}{3 - 3} = \frac{0}{0} \][/tex]
Because we get an indeterminate form [tex]\(\frac{0}{0}\)[/tex], we need to use algebraic techniques or L'Hôpital's rule to simplify this expression and find the limit.
2. Simplify the expression by factoring or rationalization:
Notice that the numerator [tex]\(4 - n^2\)[/tex] can be factored using the difference of squares:
[tex]\[ 4 - n^2 = (2 + n)(2 - n) \][/tex]
Therefore, the function becomes:
[tex]\[ \frac{(2 + n)(2 - n)}{3 - \sqrt{n^2 + 5}} \][/tex]
3. Rationalize the denominator:
To handle the square root, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator [tex]\(3 + \sqrt{n^2 + 5}\)[/tex]:
[tex]\[ \frac{(2 + n)(2 - n)}{3 - \sqrt{n^2+5}} \cdot \frac{3 + \sqrt{n^2+5}}{3 + \sqrt{n^2+5}} \][/tex]
Now, multiplying:
[tex]\[ \frac{(2 + n)(2 - n)(3 + \sqrt{n^2 + 5})}{(3 - \sqrt{n^2 + 5})(3 + \sqrt{n^2 + 5})} \][/tex]
The denominator is a difference of squares:
[tex]\[ (3 - \sqrt{n^2 + 5})(3 + \sqrt{n^2 + 5}) = 3^2 - (\sqrt{n^2+5})^2 = 9 - (n^2 + 5) = 9 - n^2 - 5 = 4 - n^2 \][/tex]
So the expression simplifies to:
[tex]\[ \frac{(2+n)(2-n)(3 + \sqrt{n^2 + 5})}{4 - n^2} \][/tex]
Since [tex]\(4 - n^2 = (2-n)(2+n)\)[/tex], the numerator and the denominator cancel out:
[tex]\[ \frac{\cancel{(2+n)(2-n)}(3 + \sqrt{n^2 + 5})}{\cancel{(2+n)(2-n)}} = 3 + \sqrt{n^2 + 5} \][/tex]
4. Substitute [tex]\( n \)[/tex] with the given value:
Now, substitute [tex]\( n = 2 \)[/tex] back into the simplified expression:
[tex]\[ 3 + \sqrt{(2)^2 + 5} = 3 + \sqrt{4 + 5} = 3 + \sqrt{9} = 3 + 3 = 6 \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{{n \to 2}} \frac{4-n^2}{3-\sqrt{n^2+5}} = 6 \][/tex]
1. Substitute the value where [tex]\( n \)[/tex] approaches into the expression:
Substitute [tex]\( n = 2 \)[/tex] into the function:
[tex]\[ \frac{4 - (2)^2}{3 - \sqrt{(2)^2 + 5}} \][/tex]
Evaluating inside the parentheses:
[tex]\[ \frac{4 - 4}{3 - \sqrt{4 + 5}} = \frac{0}{3 - \sqrt{9}} \][/tex]
Since [tex]\(\sqrt{9} = 3\)[/tex], we get:
[tex]\[ \frac{0}{3 - 3} = \frac{0}{0} \][/tex]
Because we get an indeterminate form [tex]\(\frac{0}{0}\)[/tex], we need to use algebraic techniques or L'Hôpital's rule to simplify this expression and find the limit.
2. Simplify the expression by factoring or rationalization:
Notice that the numerator [tex]\(4 - n^2\)[/tex] can be factored using the difference of squares:
[tex]\[ 4 - n^2 = (2 + n)(2 - n) \][/tex]
Therefore, the function becomes:
[tex]\[ \frac{(2 + n)(2 - n)}{3 - \sqrt{n^2 + 5}} \][/tex]
3. Rationalize the denominator:
To handle the square root, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator [tex]\(3 + \sqrt{n^2 + 5}\)[/tex]:
[tex]\[ \frac{(2 + n)(2 - n)}{3 - \sqrt{n^2+5}} \cdot \frac{3 + \sqrt{n^2+5}}{3 + \sqrt{n^2+5}} \][/tex]
Now, multiplying:
[tex]\[ \frac{(2 + n)(2 - n)(3 + \sqrt{n^2 + 5})}{(3 - \sqrt{n^2 + 5})(3 + \sqrt{n^2 + 5})} \][/tex]
The denominator is a difference of squares:
[tex]\[ (3 - \sqrt{n^2 + 5})(3 + \sqrt{n^2 + 5}) = 3^2 - (\sqrt{n^2+5})^2 = 9 - (n^2 + 5) = 9 - n^2 - 5 = 4 - n^2 \][/tex]
So the expression simplifies to:
[tex]\[ \frac{(2+n)(2-n)(3 + \sqrt{n^2 + 5})}{4 - n^2} \][/tex]
Since [tex]\(4 - n^2 = (2-n)(2+n)\)[/tex], the numerator and the denominator cancel out:
[tex]\[ \frac{\cancel{(2+n)(2-n)}(3 + \sqrt{n^2 + 5})}{\cancel{(2+n)(2-n)}} = 3 + \sqrt{n^2 + 5} \][/tex]
4. Substitute [tex]\( n \)[/tex] with the given value:
Now, substitute [tex]\( n = 2 \)[/tex] back into the simplified expression:
[tex]\[ 3 + \sqrt{(2)^2 + 5} = 3 + \sqrt{4 + 5} = 3 + \sqrt{9} = 3 + 3 = 6 \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{{n \to 2}} \frac{4-n^2}{3-\sqrt{n^2+5}} = 6 \][/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.