Answer :
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]
