use lhospitals rule to evaluate the limit. lim_x→0 (2√(x + 4)-4-(1/2)x)/x² a. -1/48 b. 1/48 c. -1/32 d. 1/32

Explanation:

Step1: Check the form

When (x\rightarrow0), the numerator (2\sqrt{x + 4}-4-\frac{1}{2}x=2\sqrt{4}-4 - 0=4 - 4=0) and the denominator (x^{2}=0). So, it is in the (\frac{0}{0}) - form, and L'Hopital's Rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of the numerator: The derivative of (y = 2\sqrt{x + 4}-4-\frac{1}{2}x) is (y^\prime=2\times\frac{1}{2\sqrt{x + 4}}-0-\frac{1}{2}=\frac{1}{\sqrt{x + 4}}-\frac{1}{2}). The derivative of the denominator (y = x^{2}) is (y^\prime = 2x). So, the new limit is (\lim_{x\rightarrow0}\frac{\frac{1}{\sqrt{x + 4}}-\frac{1}{2}}{2x}).

Step3: Check the new - form

When (x\rightarrow0), the numerator (\frac{1}{\sqrt{0 + 4}}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0) and the denominator (2x = 0). It is still in the (\frac{0}{0}) - form, so apply L'Hopital's Rule again.

Step4: Differentiate numerator and denominator again

The derivative of the numerator: The derivative of (y=\frac{1}{\sqrt{x + 4}}-\frac{1}{2}=(x + 4)^{-\frac{1}{2}}-\frac{1}{2}), using the power - rule ((u^{n})^\prime=nu^{n - 1}u^\prime), we have (y^\prime=-\frac{1}{2}(x + 4)^{-\frac{3}{2}}). The derivative of the denominator (y = 2x) is (y^\prime=2). So, the new limit is (\lim_{x\rightarrow0}\frac{-\frac{1}{2}(x + 4)^{-\frac{3}{2}}}{2}).

Step5: Evaluate the limit

Substitute (x = 0) into (\frac{-\frac{1}{2}(x + 4)^{-\frac{3}{2}}}{2}), we get (\frac{-\frac{1}{2}(0 + 4)^{-\frac{3}{2}}}{2}=\frac{-\frac{1}{2}\times\frac{1}{8}}{2}=-\frac{1}{32}).

Answer:

C. (-\frac{1}{32})