Question

The value of \[ \lim_{x \to 1} \frac{x^4 - \sqrt{x}}{\sqrt{x} - 1} \] is:

• A. \( 7 \)
• B. does not exist
• C. \( 1 \)
• D. \( 0 \)

Hint:

L'Hopital's Rule can be applied to evaluate limits that result in indeterminate forms such as \( \frac{0}{0} \).

Answer 1

The Correct Option is A

We need to evaluate the limit: \[ \lim_{x \to 1} \frac{x^4 - \sqrt{x}}{\sqrt{x} - 1} \] Using L'Hopital's Rule (since the limit yields a \( \frac{0}{0} \) indeterminate form), we differentiate the numerator and denominator: The numerator is \( \frac{d}{dx}(x^4 - \sqrt{x}) = 4x^3 - \frac{1}{2\sqrt{x}} \). The denominator is \( \frac{d}{dx}(\sqrt{x} - 1) = \frac{1}{2\sqrt{x}} \). Now, applying L'Hopital's Rule: \[ \lim_{x \to 1} \frac{4x^3 - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} \] Substitute \( x = 1 \): \[ \frac{4(1)^3 - \frac{1}{2}}{\frac{1}{2}} = \frac{4 - \frac{1}{2}}{\frac{1}{2}} = \frac{\frac{7}{2}}{\frac{1}{2}} = 7 \] Thus, the value of the limit is \( 7 \).