Answer :
Sure, let's solve the limit problem:
[tex]\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - \sqrt{x^2 - x - 1} \right) \][/tex]
To solve this, we will first simplify the expression inside the limit.
Notice that the terms inside the square roots, [tex]\( x^2 + x + 1 \)[/tex] and [tex]\( x^2 - x - 1 \)[/tex], both have [tex]\( x^2 \)[/tex] which dominates as [tex]\( x \)[/tex] increases to infinity.
1. Factor out [tex]\( x^2 \)[/tex] from both square roots:
[tex]\[ \sqrt{x^2 + x + 1} = \sqrt{x^2\left( 1 + \frac{x}{x^2} + \frac{1}{x^2} \right)} = \sqrt{x^2 \left( 1 + \frac{1}{x} + \frac{1}{x^2} \right)} \][/tex]
[tex]\[ \sqrt{x^2 + x + 1} = x \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} \][/tex]
Similarly,
[tex]\[ \sqrt{x^2 - x - 1} = \sqrt{x^2\left( 1 - \frac{x}{x^2} - \frac{1}{x^2} \right)} = \sqrt{x^2 \left( 1 - \frac{1}{x} - \frac{1}{x^2} \right)} \][/tex]
[tex]\[ \sqrt{x^2 - x - 1} = x \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \][/tex]
2. Rewrite the original limit with these simplified expressions:
[tex]\[ \lim_{x \to \infty} \left( x \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - x \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \right) \][/tex]
3. Factor out [tex]\( x \)[/tex]:
[tex]\[ x \left( \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \right) \][/tex]
4. Now consider the behavior of the terms inside the square roots as [tex]\( x \)[/tex] goes to infinity. As [tex]\( x \)[/tex] approaches infinity, [tex]\( \frac{1}{x} \)[/tex] approaches 0 and [tex]\( \frac{1}{x^2} \)[/tex] also approaches 0. Therefore, the terms inside the square roots approach 1.
Let's use the binomial expansion for square roots for large [tex]\( x \)[/tex]:
[tex]\[ \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} \approx 1 + \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \][/tex]
[tex]\[ \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \approx 1 - \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \][/tex]
5. Subtract these two expressions:
[tex]\[ \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \approx \left( 1 + \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \right) - \left( 1 - \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \right) \][/tex]
[tex]\[ \approx \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) + \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \][/tex]
[tex]\[ \approx \frac{1}{x} + \frac{1}{x^2} \][/tex]
6. Multiply by [tex]\( x \)[/tex]:
[tex]\[ x \left( \frac{1}{x} + \frac{1}{x^2} \right) = 1 + \frac{1}{x} \][/tex]
As [tex]\( x \)[/tex] goes to infinity, [tex]\( \frac{1}{x} \)[/tex] approaches 0. Therefore, the expression approaches 1.
So, the answer is:
[tex]\[ \boxed{1} \][/tex]
[tex]\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - \sqrt{x^2 - x - 1} \right) \][/tex]
To solve this, we will first simplify the expression inside the limit.
Notice that the terms inside the square roots, [tex]\( x^2 + x + 1 \)[/tex] and [tex]\( x^2 - x - 1 \)[/tex], both have [tex]\( x^2 \)[/tex] which dominates as [tex]\( x \)[/tex] increases to infinity.
1. Factor out [tex]\( x^2 \)[/tex] from both square roots:
[tex]\[ \sqrt{x^2 + x + 1} = \sqrt{x^2\left( 1 + \frac{x}{x^2} + \frac{1}{x^2} \right)} = \sqrt{x^2 \left( 1 + \frac{1}{x} + \frac{1}{x^2} \right)} \][/tex]
[tex]\[ \sqrt{x^2 + x + 1} = x \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} \][/tex]
Similarly,
[tex]\[ \sqrt{x^2 - x - 1} = \sqrt{x^2\left( 1 - \frac{x}{x^2} - \frac{1}{x^2} \right)} = \sqrt{x^2 \left( 1 - \frac{1}{x} - \frac{1}{x^2} \right)} \][/tex]
[tex]\[ \sqrt{x^2 - x - 1} = x \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \][/tex]
2. Rewrite the original limit with these simplified expressions:
[tex]\[ \lim_{x \to \infty} \left( x \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - x \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \right) \][/tex]
3. Factor out [tex]\( x \)[/tex]:
[tex]\[ x \left( \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \right) \][/tex]
4. Now consider the behavior of the terms inside the square roots as [tex]\( x \)[/tex] goes to infinity. As [tex]\( x \)[/tex] approaches infinity, [tex]\( \frac{1}{x} \)[/tex] approaches 0 and [tex]\( \frac{1}{x^2} \)[/tex] also approaches 0. Therefore, the terms inside the square roots approach 1.
Let's use the binomial expansion for square roots for large [tex]\( x \)[/tex]:
[tex]\[ \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} \approx 1 + \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \][/tex]
[tex]\[ \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \approx 1 - \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \][/tex]
5. Subtract these two expressions:
[tex]\[ \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}} \approx \left( 1 + \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \right) - \left( 1 - \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \right) \][/tex]
[tex]\[ \approx \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) + \frac{1}{2} \left( \frac{1}{x} + \frac{1}{x^2} \right) \][/tex]
[tex]\[ \approx \frac{1}{x} + \frac{1}{x^2} \][/tex]
6. Multiply by [tex]\( x \)[/tex]:
[tex]\[ x \left( \frac{1}{x} + \frac{1}{x^2} \right) = 1 + \frac{1}{x} \][/tex]
As [tex]\( x \)[/tex] goes to infinity, [tex]\( \frac{1}{x} \)[/tex] approaches 0. Therefore, the expression approaches 1.
So, the answer is:
[tex]\[ \boxed{1} \][/tex]
