Answer :
To solve the limit [tex]\(\lim _{x \rightarrow a} \frac{\sqrt{x+a}-\sqrt{3 x-a}}{x-a}\)[/tex], we will use the process of finding the limit by algebraic manipulation and applying limit properties. Here is a detailed, step-by-step solution for the given limit:
1. Consider the given limit:
[tex]\[ \lim _{x \rightarrow a} \frac{\sqrt{x+a}-\sqrt{3 x-a}}{x-a} \][/tex]
2. Identify the indeterminate form:
Direct substitution of [tex]\( x = a \)[/tex] in the expression [tex]\(\frac{\sqrt{x+a}-\sqrt{3 x-a}}{x-a}\)[/tex] leads to:
[tex]\[ \frac{\sqrt{a+a}-\sqrt{3a-a}}{a-a} = \frac{\sqrt{2a}-\sqrt{2a}}{0} = \frac{0}{0} \][/tex]
This is an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex].
3. Rationalize the numerator:
To eliminate the indeterminate form, multiply the numerator and the denominator by the conjugate of the numerator:
[tex]\[ \frac{\sqrt{x+a}-\sqrt{3x-a}}{x-a} \cdot \frac{\sqrt{x+a}+\sqrt{3x-a}}{\sqrt{x+a}+\sqrt{3x-a}} \][/tex]
This gives us:
[tex]\[ \frac{(\sqrt{x+a}-\sqrt{3x-a})(\sqrt{x+a}+\sqrt{3x-a})}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
4. Simplify the expression:
The numerator simplifies using the difference of squares formula [tex]\((a-b)(a+b) = a^2 - b^2\)[/tex]:
[tex]\[ \frac{(\sqrt{x+a})^2-(\sqrt{3x-a})^2}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
[tex]\[ = \frac{(x+a) - (3x-a)}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
[tex]\[ = \frac{x + a - 3x + a}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
[tex]\[ = \frac{2a - 2x}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
5. Factor out common terms in the numerator:
Notice that [tex]\(2a - 2x = -2(x - a)\)[/tex]:
[tex]\[ = \frac{-2(x-a)}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
6. Cancel the common factors:
[tex]\[ = \frac{-2}{\sqrt{x+a}+\sqrt{3x-a}} \][/tex]
7. Evaluate the limit as [tex]\( x \)[/tex] approaches [tex]\( a \)[/tex]:
Substitute [tex]\( x = a \)[/tex] into the simplified expression in the denominator:
[tex]\[ \lim_{x \to a} \frac{-2}{\sqrt{x+a}+\sqrt{3x-a}} = \frac{-2}{\sqrt{a+a}+\sqrt{3a-a}} \][/tex]
[tex]\[ = \frac{-2}{\sqrt{2a}+\sqrt{2a}} = \frac{-2}{2\sqrt{2a}} \][/tex]
8. Simplify the final result:
[tex]\[ = \frac{-1}{\sqrt{2a}} = -\frac{1}{\sqrt{2a}} \][/tex]
9. Rationalize the denominator (if required):
Sometimes we need to rewrite the result in a standardized form. Here, the result can also be written as:
[tex]\[ -\frac{\sqrt{2}}{2\sqrt{a}} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{-\frac{\sqrt{2}}{2\sqrt{a}}} \][/tex]
1. Consider the given limit:
[tex]\[ \lim _{x \rightarrow a} \frac{\sqrt{x+a}-\sqrt{3 x-a}}{x-a} \][/tex]
2. Identify the indeterminate form:
Direct substitution of [tex]\( x = a \)[/tex] in the expression [tex]\(\frac{\sqrt{x+a}-\sqrt{3 x-a}}{x-a}\)[/tex] leads to:
[tex]\[ \frac{\sqrt{a+a}-\sqrt{3a-a}}{a-a} = \frac{\sqrt{2a}-\sqrt{2a}}{0} = \frac{0}{0} \][/tex]
This is an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex].
3. Rationalize the numerator:
To eliminate the indeterminate form, multiply the numerator and the denominator by the conjugate of the numerator:
[tex]\[ \frac{\sqrt{x+a}-\sqrt{3x-a}}{x-a} \cdot \frac{\sqrt{x+a}+\sqrt{3x-a}}{\sqrt{x+a}+\sqrt{3x-a}} \][/tex]
This gives us:
[tex]\[ \frac{(\sqrt{x+a}-\sqrt{3x-a})(\sqrt{x+a}+\sqrt{3x-a})}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
4. Simplify the expression:
The numerator simplifies using the difference of squares formula [tex]\((a-b)(a+b) = a^2 - b^2\)[/tex]:
[tex]\[ \frac{(\sqrt{x+a})^2-(\sqrt{3x-a})^2}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
[tex]\[ = \frac{(x+a) - (3x-a)}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
[tex]\[ = \frac{x + a - 3x + a}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
[tex]\[ = \frac{2a - 2x}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
5. Factor out common terms in the numerator:
Notice that [tex]\(2a - 2x = -2(x - a)\)[/tex]:
[tex]\[ = \frac{-2(x-a)}{(x-a)(\sqrt{x+a}+\sqrt{3x-a})} \][/tex]
6. Cancel the common factors:
[tex]\[ = \frac{-2}{\sqrt{x+a}+\sqrt{3x-a}} \][/tex]
7. Evaluate the limit as [tex]\( x \)[/tex] approaches [tex]\( a \)[/tex]:
Substitute [tex]\( x = a \)[/tex] into the simplified expression in the denominator:
[tex]\[ \lim_{x \to a} \frac{-2}{\sqrt{x+a}+\sqrt{3x-a}} = \frac{-2}{\sqrt{a+a}+\sqrt{3a-a}} \][/tex]
[tex]\[ = \frac{-2}{\sqrt{2a}+\sqrt{2a}} = \frac{-2}{2\sqrt{2a}} \][/tex]
8. Simplify the final result:
[tex]\[ = \frac{-1}{\sqrt{2a}} = -\frac{1}{\sqrt{2a}} \][/tex]
9. Rationalize the denominator (if required):
Sometimes we need to rewrite the result in a standardized form. Here, the result can also be written as:
[tex]\[ -\frac{\sqrt{2}}{2\sqrt{a}} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{-\frac{\sqrt{2}}{2\sqrt{a}}} \][/tex]
