Certainly! Let's evaluate each limit step-by-step.
### Part (a) Evaluation:
Evaluate:
[tex]\[ \lim _{x \rightarrow 25} \frac{x-25}{\sqrt{x}-5} \][/tex]
Step-by-Step Solution:
1. Identify the form: When substituting [tex]\( x = 25 \)[/tex], the numerator becomes [tex]\(25 - 25 = 0\)[/tex] and the denominator becomes [tex]\( \sqrt{25} - 5 = 0 \)[/tex]. This is an indeterminate form [tex]\( \frac{0}{0} \)[/tex].
2. Simplify the expression: Notice that we can use a factorization technique, specifically the difference of squares:
[tex]\[ x - 25 = (\sqrt{x})^2 - 5^2 \][/tex]
3. Rewrite the expression:
[tex]\[ \frac{x-25}{\sqrt{x}-5} = \frac{(\sqrt{x})^2 - 5^2}{\sqrt{x}-5} \][/tex]
4. Factor the numerator:
[tex]\[ \frac{(\sqrt{x}-5)(\sqrt{x}+5)}{\sqrt{x}-5} \][/tex]
5. Simplify the fraction by canceling out the common terms:
[tex]\[ = \sqrt{x} + 5 \][/tex]
6. Evaluate the limit: Substitute [tex]\( x = 25 \)[/tex]:
[tex]\[ \sqrt{25} + 5 = 5 + 5 = 10 \][/tex]
So, the limit is:
[tex]\[ \lim _{x \rightarrow 25} \frac{x-25}{\sqrt{x}-5} = 10 \][/tex]
### Part (c) Evaluation:
Evaluate:
[tex]\[ \lim _{x \rightarrow 1} \frac{\sqrt{3x+1}-2}{x-1} \][/tex]
Step-by-Step Solution:
1. Identify the form: When substituting [tex]\( x = 1 \)[/tex], the numerator becomes [tex]\( \sqrt{3 \cdot 1 + 1} - 2 = \sqrt{4} - 2 = 2 - 2 = 0 \)[/tex] and the denominator becomes [tex]\( 1 - 1 = 0 \)[/tex]. This is an indeterminate form [tex]\( \frac{0}{0} \)[/tex].
2. Apply L'Hopital's Rule: L'Hopital's Rule states that if the limit results in an indeterminate form, we can differentiate the numerator and the denominator separately and then take the limit again.
3. Differentiate the numerator:
Let [tex]\( f(x) = \sqrt{3x + 1} \)[/tex].
[tex]\[ f'(x) = \frac{d}{dx} (\sqrt{3x + 1}) = \frac{1}{2\sqrt{3x+1}} \cdot 3 = \frac{3}{2\sqrt{3x+1}} \][/tex]
4. Differentiate the denominator:
Let [tex]\( g(x) = x - 1 \)[/tex].
[tex]\[ g'(x) = \frac{d}{dx} (x - 1) = 1 \][/tex]
5. Reevaluate the limit with the derivatives:
[tex]\[ \lim_{x \rightarrow 1} \frac{\frac{3}{2\sqrt{3x + 1}}}{1} = \lim_{x \rightarrow 1} \frac{3}{2\sqrt{3x + 1}} \][/tex]
6. Substitute [tex]\( x = 1 \)[/tex] into the simplified expression:
[tex]\[ \frac{3}{2\sqrt{3 \cdot 1 + 1}} = \frac{3}{2\sqrt{4}} = \frac{3}{2 \cdot 2} = \frac{3}{4} \][/tex]
So, the limit is:
[tex]\[ \lim _{x \rightarrow 1} \frac{\sqrt{3x + 1} - 2}{x - 1} = \frac{3}{4} \][/tex]
### Final Results:
(a) [tex]\(\lim _{x \rightarrow 25} \frac{x-25}{\sqrt{x}-5} = 10\)[/tex]
(c) [tex]\(\lim _{x \rightarrow 1} \frac{\sqrt{3x + 1}-2}{x-1} = \frac{3}{4}\)[/tex]
