Schofield Reed
02/26/2024 · Senior High School
\( \lim _{x \rightarrow 0} \frac{\sqrt{a x+b}-2}{x}=1 \) find \( a \) and \( b \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the limit
\[
\lim_{x \rightarrow 0} \frac{\sqrt{ax + b} - 2}{x} = 1,
\]
we start by substituting \( x = 0 \) into the expression inside the limit. This gives us:
\[
\sqrt{a(0) + b} - 2 = \sqrt{b} - 2.
\]
For the limit to be defined, we need \( \sqrt{b} - 2 = 0 \), which implies:
\[
\sqrt{b} = 2 \quad \Rightarrow \quad b = 4.
\]
Now, substituting \( b = 4 \) into the limit, we have:
\[
\lim_{x \rightarrow 0} \frac{\sqrt{ax + 4} - 2}{x}.
\]
Next, we can apply L'Hôpital's Rule since the limit is in the indeterminate form \( \frac{0}{0} \). We differentiate the numerator and the denominator:
1. The derivative of the numerator \( \sqrt{ax + 4} - 2 \) is:
\[
\frac{d}{dx}(\sqrt{ax + 4}) = \frac{1}{2\sqrt{ax + 4}} \cdot a = \frac{a}{2\sqrt{ax + 4}}.
\]
2. The derivative of the denominator \( x \) is \( 1 \).
Applying L'Hôpital's Rule, we get:
\[
\lim_{x \rightarrow 0} \frac{\sqrt{ax + 4} - 2}{x} = \lim_{x \rightarrow 0} \frac{\frac{a}{2\sqrt{ax + 4}}}{1} = \frac{a}{2\sqrt{4}} = \frac{a}{4}.
\]
Setting this equal to 1 (as given in the problem), we have:
\[
\frac{a}{4} = 1 \quad \Rightarrow \quad a = 4.
\]
Thus, the values of \( a \) and \( b \) are:
\[
\boxed{(4, 4)}.
\]
Quick Answer
\( a = 4 \) and \( b = 4 \).
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