Slope of a tangent = limit of the function?

nejnadusho
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Slope of a tangent = limit of the function?

Hi.


I want to ask .
Can it be true that

The slope of a tangent line is a limit of a function.

or

The slope of a tangent line can be considered as a limit of a function?

?

Is it the same I mean the slope of a tangent line and a limit of a function
or I am dreaming something?

thanks in advance
 
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Did you mean the limit of the rate of change of a function? Maybe you should explain more what you meant by "the limit"
 
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The slope of a line tangent to a function f at a point x is the derivative of that function f, evaluated at x. The derivative of a function at a fixed point x is, by definition, a limit. Do you know the definition of f'(x)?
 
The definition?
I am not sure.

I know that this is a function. And depends what kind of function is if I imput a value for
x I will have an aoutput.

And also for a sertain value of x I will receive a certain value for y.
 
Has your professor/teacher not gone over derivatives in class yet? If not, the Wikipedia entry on derivatives will give you the definition (and the solution!).
 
Thank you guys once again.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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