Limit Evaluation: Understanding the Indeterminate Form

The question asks us to evaluate the limit of the expression \(\sqrt{x^2 + x - 1} - x\) as \(x\) approaches infinity. This is a common type of limit problem encountered in calculus.

The given limit is:

\[ \mathop {\lim }\limits_{x \to \infty } \left( \sqrt {{x^2} + x - 1} - x \right) \]

When \(x \to \infty\), the term \(\sqrt{x^2 + x - 1}\) approaches \(\infty\) and the term \(x\) also approaches \(\infty\). This leads to an indeterminate form of type \(\infty - \infty\). To resolve this, we typically use the method of multiplying by the conjugate.

Conjugate Multiplication for Limit Simplification

To eliminate the indeterminate form, we multiply and divide the expression by its conjugate. The conjugate of \(\sqrt{A} - B\) is \(\sqrt{A} + B\). In our case, \(A = x^2 + x - 1\) and \(B = x\), so the conjugate is \(\sqrt{x^2 + x - 1} + x\).

Let's perform the multiplication:

\[ \mathop {\lim }\limits_{x \to \infty } \left( \sqrt {{x^2} + x - 1} - x \right) \times \frac{\sqrt {{x^2} + x - 1} + x}{\sqrt {{x^2} + x - 1} + x} \]

Using the algebraic identity \((a - b)(a + b) = a^2 - b^2\), where \(a = \sqrt{x^2 + x - 1}\) and \(b = x\), the numerator simplifies as follows:

\[ = \mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt {{x^2} + x - 1})^2 - x^2}{\sqrt {{x^2} + x - 1} + x} \]

\[ = \mathop {\lim }\limits_{x \to \infty } \frac{(x^2 + x - 1) - x^2}{\sqrt {{x^2} + x - 1} + x} \]

\[ = \mathop {\lim }\limits_{x \to \infty } \frac{x - 1}{\sqrt {{x^2} + x - 1} + x} \]

Simplifying the Denominator for Limit Evaluation

Now we have a rational expression. To evaluate the limit as \(x \to \infty\), we need to divide both the numerator and the denominator by the highest power of \(x\) in the denominator. Let's first simplify the term inside the square root in the denominator.

For large positive \(x\), we can factor \(x^2\) out of the square root:

\[ \sqrt{x^2 + x - 1} = \sqrt{x^2 \left( 1 + \frac{x}{x^2} - \frac{1}{x^2} \right)} = \sqrt{x^2 \left( 1 + \frac{1}{x} - \frac{1}{x^2} \right)} \]

Since \(x \to \infty\), \(x\) is positive, so \(\sqrt{x^2} = |x| = x\).

\[ = x \sqrt{1 + \frac{1}{x} - \frac{1}{x^2}} \]

Substitute this back into our limit expression:

\[ = \mathop {\lim }\limits_{x \to \infty } \frac{x - 1}{x \sqrt{1 + \frac{1}{x} - \frac{1}{x^2}} + x} \]

Now, divide both the numerator and the denominator by \(x\):

\[ = \mathop {\lim }\limits_{x \to \infty } \frac{\frac{x}{x} - \frac{1}{x}}{\frac{x \sqrt{1 + \frac{1}{x} - \frac{1}{x^2}}}{x} + \frac{x}{x}} \]

\[ = \mathop {\lim }\limits_{x \to \infty } \frac{1 - \frac{1}{x}}{\sqrt{1 + \frac{1}{x} - \frac{1}{x^2}} + 1} \]

Evaluating the Limit as x Approaches Infinity

As \(x \to \infty\), we know that terms of the form \(\frac{c}{x^n}\) (where \(c\) is a constant and \(n > 0\)) approach \(0\).

• \(\frac{1}{x} \to 0\)
• \(\frac{1}{x^2} \to 0\)

Substituting these values into the simplified limit expression:

\[ = \frac{1 - 0}{\sqrt{1 + 0 - 0} + 1} \]

\[ = \frac{1}{\sqrt{1} + 1} \]

\[ = \frac{1}{1 + 1} \]

\[ = \frac{1}{2} \]

Therefore, the limit of the given expression is \(\frac{1}{2}\).