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Chapter 2: Q. 2TF (page 168)

Instead of choosing small values of h, we could have chosen values of z close to c. What limit involving z instead of h is equivalent to the one involving h?

Short Answer

Expert verified

The derivative value isf'(c)=limz→cf(z)-f(c)z-c.

Step by step solution

01

Step 1.Given Information

Given as h tends to zero, z close to c

02

Step 2.Forming limits 

If the limit of these quantities approaches a real number as h→0, or as z→c, then we

will define that real number to be the derivative of f at the point x=c.

03

Step 3.The solution 

As we know that z=c+has h→0;z→c.The value of h=z-c.

The derivative value isf'(c)=limh→0f(c+h)-f(c)hf'(c)=limz→cf(z)-f(c)z-c.

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