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This may be a dumb question, but I'll ask anyway...
Find the values of a and b such that
\lim_{x \rightarrow 0} \frac{\sqrt{a + bx} - \sqrt{3}}{x} = \sqrt{3}
N/A
I already have the work and the solution. However, someone showed me a different way. Here are the first couple of steps:
\lim_{x \rightarrow 0} \frac{\sqrt{a + bx} - \sqrt{3}}{x} = \sqrt{3}
\lim_{x \rightarrow 0} \left( \sqrt{a + bx} - \sqrt{3} \right) = \sqrt{3} \cdot x
\lim_{x \rightarrow 0} \sqrt{a + bx} = \sqrt{3} + \sqrt{3} \cdot x
I'm having a brain fart. Is multiplying both sides by x, and then adding sqrt 3 to both sides, legal?
Homework Statement
Find the values of a and b such that
\lim_{x \rightarrow 0} \frac{\sqrt{a + bx} - \sqrt{3}}{x} = \sqrt{3}
Homework Equations
N/A
The Attempt at a Solution
I already have the work and the solution. However, someone showed me a different way. Here are the first couple of steps:
\lim_{x \rightarrow 0} \frac{\sqrt{a + bx} - \sqrt{3}}{x} = \sqrt{3}
\lim_{x \rightarrow 0} \left( \sqrt{a + bx} - \sqrt{3} \right) = \sqrt{3} \cdot x
\lim_{x \rightarrow 0} \sqrt{a + bx} = \sqrt{3} + \sqrt{3} \cdot x
I'm having a brain fart. Is multiplying both sides by x, and then adding sqrt 3 to both sides, legal?