Answer :
Certainly! Let's evaluate the limit step by step:
We want to find:
[tex]\[ \lim_{x \rightarrow \infty} \sqrt{x}(\sqrt{x+3} - \sqrt{x}) \][/tex]
1. Rewrite the expression:
First, note that as [tex]\( x \)[/tex] grows very large, both [tex]\( \sqrt{x+3} \)[/tex] and [tex]\( \sqrt{x} \)[/tex] will also become very large, and their difference might become small. To handle this, let's simplify the expression by rationalizing the term inside the parentheses:
[tex]\[ \sqrt{x+3} - \sqrt{x} = \frac{(\sqrt{x+3} - \sqrt{x})(\sqrt{x+3} + \sqrt{x})}{\sqrt{x+3} + \sqrt{x}} \][/tex]
This simplifies to:
[tex]\[ \sqrt{x+3} - \sqrt{x} = \frac{(x+3) - x}{\sqrt{x+3} + \sqrt{x}} = \frac{3}{\sqrt{x+3} + \sqrt{x}} \][/tex]
Substituting this back into the original expression, we get:
[tex]\[ \sqrt{x} \left( \frac{3}{\sqrt{x+3} + \sqrt{x}} \right) = \frac{3\sqrt{x}}{\sqrt{x+3} + \sqrt{x}} \][/tex]
2. Simplify the new expression:
Next, let's handle the expression under the limit separately:
[tex]\[ \frac{3\sqrt{x}}{\sqrt{x+3} + \sqrt{x}} \][/tex]
As [tex]\( x \)[/tex] approaches infinity, [tex]\( \sqrt{x+3} \)[/tex] approaches [tex]\( \sqrt{x} \)[/tex] because the extra 3 inside the square root becomes negligible. So:
[tex]\[ \sqrt{x+3} \approx \sqrt{x} \][/tex]
Thus, the denominator simplifies to:
[tex]\[ \sqrt{x+3} + \sqrt{x} \approx \sqrt{x} + \sqrt{x} = 2\sqrt{x} \][/tex]
So the expression becomes:
[tex]\[ \frac{3\sqrt{x}}{2\sqrt{x}} = \frac{3}{2} \][/tex]
3. Taking the limit:
At this point, we can see that:
[tex]\[ \lim_{x \rightarrow \infty} \frac{3\sqrt{x}}{\sqrt{x+3} + \sqrt{x}} = \frac{3}{2} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{\frac{3}{2}} \][/tex]
We want to find:
[tex]\[ \lim_{x \rightarrow \infty} \sqrt{x}(\sqrt{x+3} - \sqrt{x}) \][/tex]
1. Rewrite the expression:
First, note that as [tex]\( x \)[/tex] grows very large, both [tex]\( \sqrt{x+3} \)[/tex] and [tex]\( \sqrt{x} \)[/tex] will also become very large, and their difference might become small. To handle this, let's simplify the expression by rationalizing the term inside the parentheses:
[tex]\[ \sqrt{x+3} - \sqrt{x} = \frac{(\sqrt{x+3} - \sqrt{x})(\sqrt{x+3} + \sqrt{x})}{\sqrt{x+3} + \sqrt{x}} \][/tex]
This simplifies to:
[tex]\[ \sqrt{x+3} - \sqrt{x} = \frac{(x+3) - x}{\sqrt{x+3} + \sqrt{x}} = \frac{3}{\sqrt{x+3} + \sqrt{x}} \][/tex]
Substituting this back into the original expression, we get:
[tex]\[ \sqrt{x} \left( \frac{3}{\sqrt{x+3} + \sqrt{x}} \right) = \frac{3\sqrt{x}}{\sqrt{x+3} + \sqrt{x}} \][/tex]
2. Simplify the new expression:
Next, let's handle the expression under the limit separately:
[tex]\[ \frac{3\sqrt{x}}{\sqrt{x+3} + \sqrt{x}} \][/tex]
As [tex]\( x \)[/tex] approaches infinity, [tex]\( \sqrt{x+3} \)[/tex] approaches [tex]\( \sqrt{x} \)[/tex] because the extra 3 inside the square root becomes negligible. So:
[tex]\[ \sqrt{x+3} \approx \sqrt{x} \][/tex]
Thus, the denominator simplifies to:
[tex]\[ \sqrt{x+3} + \sqrt{x} \approx \sqrt{x} + \sqrt{x} = 2\sqrt{x} \][/tex]
So the expression becomes:
[tex]\[ \frac{3\sqrt{x}}{2\sqrt{x}} = \frac{3}{2} \][/tex]
3. Taking the limit:
At this point, we can see that:
[tex]\[ \lim_{x \rightarrow \infty} \frac{3\sqrt{x}}{\sqrt{x+3} + \sqrt{x}} = \frac{3}{2} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{\frac{3}{2}} \][/tex]
