What does it tell us and how could it be used in differentiation?

Asked Jun 13, 2014 by anonymous

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Differentiation only works when the curve exists from both sides, otherwise you can't really speak of a "gradient" (which is essentially what differentiation is aiming to achieve). This is why you need to find a limit - which is essentially a calculation done by approaching it from one side or the other (e.g. because you can't divide by zero, let's see what happens when we divide by 0.1, then 0.01, then 0.001, and see what it converges to?) The example in the brackets above explains why the limit is usually h=0, because you can avoid dividing by zero with a limit - which is exactly what you are doing with differentiation. You are calculating a gradient of a single point, which is the same as taking a normal straight line gradient between two points, except you are shrinking the "run" part of "rise"/"run" to zero (dividing by zero). A limit allows you to realistically achieve that calculation without dividing by zero. The gradient function reads out the gradient of any point of the original function. e.g. y=x^2, dy/dx=2x. At x=1, the gradient of y=x^2 is dy/dx=2*1=2

Differentiation only works when the curve exists from both sides, otherwise you can't really speak of a "gradient" (which is essentially what differentiation is aiming to achieve). This is why you need to find a limit - which is essentially a calculation done by approaching it from one side or the other (e.g. because you can't divide by zero, let's see what happens when we divide by 0.1, then 0.01, then 0.001, and see what it converges to?)

The example in the brackets above explains why the limit is usually h=0, because you can avoid dividing by zero with a limit - which is exactly what you are doing with differentiation. You are calculating a gradient of a single point, which is the same as taking a normal straight line gradient between two points, except you are shrinking the "run" part of "rise"/"run" to zero (dividing by zero). A limit allows you to realistically achieve that calculation without dividing by zero.

The gradient function reads out the gradient of any point of the original function. e.g. y=x^2, dy/dx=2x. At x=1, the gradient of y=x^2 is dy/dx=2*1=2

Source: http://merspi.com.au/55814/why-do-we-need-to-find-if-a-limit-exists-when-differentiating

In essence, the limit underlies the process of differentiation, but it is actually invisible in "everyday differentiation" because you just use rules such as x^n becomes nx^(n-1). If you do it by first principles, then you need to take a limit of h=0 to find an instantaneous gradient instead of an interval gradient (rise over run)

Answered Jun 13, 2014 by Collin Li (56,060 points) selected Jun 14, 2014 by Community

## Purpose of limits in differentiation?

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