# How do you find f'(x) using the limit definition given #sqrt(2x-1)#?

next, we times by the conjugate

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To find ( f'(x) ) using the limit definition for the function ( f(x) = \sqrt{2x - 1} ), you use the formula:

[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

Substitute ( f(x) = \sqrt{2x - 1} ) into the formula:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x + h) - 1} - \sqrt{2x - 1}}{h} ]

Then simplify the expression by rationalizing the numerator:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x + h) - 1} - \sqrt{2x - 1}}{h} \times \frac{\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}{\sqrt{2(x + h) - 1} + \sqrt{2x - 1}} ]

[ f'(x) = \lim_{h \to 0} \frac{2(x + h) - 1 - (2x - 1)}{h (\sqrt{2(x + h) - 1} + \sqrt{2x - 1})} ]

[ f'(x) = \lim_{h \to 0} \frac{2x + 2h - 1 - 2x + 1}{h (\sqrt{2(x + h) - 1} + \sqrt{2x - 1})} ]

[ f'(x) = \lim_{h \to 0} \frac{2h}{h (\sqrt{2(x + h) - 1} + \sqrt{2x - 1})} ]

[ f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2(x + h) - 1} + \sqrt{2x - 1}} ]

[ f'(x) = \frac{2}{\sqrt{2x - 1} + \sqrt{2x - 1}} ]

[ f'(x) = \frac{2}{2\sqrt{2x - 1}} ]

[ f'(x) = \frac{1}{\sqrt{2x - 1}} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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