How do you find the derivative of #f(x)=2x^2+x1# using the limit process?
Based on the limit definition of derivative:
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To find the derivative of (f(x) = 2x^2 + x  1) using the limit process, you can follow these steps:

Start with the definition of the derivative: [f'(x) = \lim_{h \to 0} \frac{f(x + h)  f(x)}{h}]

Substitute the given function (f(x) = 2x^2 + x  1) into the definition: [f'(x) = \lim_{h \to 0} \frac{(2(x+h)^2 + (x+h)  1)  (2x^2 + x  1)}{h}]

Expand and simplify the expression inside the limit: [f'(x) = \lim_{h \to 0} \frac{2(x^2 + 2xh + h^2) + (x + h)  1  2x^2  x + 1}{h}] [= \lim_{h \to 0} \frac{2x^2 + 4xh + 2h^2 + x + h  1  2x^2  x + 1}{h}] [= \lim_{h \to 0} \frac{4xh + 2h^2 + h}{h}]

Factor out (h) from the numerator and cancel it with (h) in the denominator: [f'(x) = \lim_{h \to 0} \frac{h(4x + 2h + 1)}{h}]

Simplify the expression: [f'(x) = \lim_{h \to 0} (4x + 2h + 1)]

Now, take the limit as (h) approaches 0: [f'(x) = 4x + 1]
So, the derivative of (f(x) = 2x^2 + x  1) is (f'(x) = 4x + 1).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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