What is \lim_{x\rightarrow0}\frac{\sqrt{f(x)}-3}{\sqrt{f(x)+7}-4} equal to?
Consider the following for the two (02) items that follow : Let the function f(x)=x^{2}+9.
- A. 2/3
- B. 1
- C. 4/3 ✓
- D. 2
Correct Answer: C. 4/3
Explanation
The limit is \lim_{x\rightarrow0}\frac{\sqrt{x^2+9}-3}{\sqrt{x^2+16}-4}, which is a 0/0 form. Multiplying numerator and denominator by their conjugates gives \frac{(x^2+9-9)(\sqrt{x^2+16}+4)}{(x^2+16-16)(\sqrt{x^2+9}+3)} = \frac{x^2(\sqrt{x^2+16}+4)}{x^2(\sqrt{x^2+9}+3)}. Cancelling x^2 and substituting x=0 yields \frac{\sqrt{16}+4}{\sqrt{9}+3} = \frac{4+4}{3+3} = \frac{8}{6} = 4/3.
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