suppose lim x-> 0 [g(x)-g(0)]/x = 1 it follows that ...

Question

a.) g is not defined at x = 0

b.) g is not continuous at x = 0

c.) The limit of g(x) as x approaches 0 equals 1

d.) 𝑔 ′ (0) = 1

e.) 𝑔 ′ (1) = 0

I'm not sure which answer is correct, i'm tied between answer c and d. I believe we're given an avg. rate of change, but i'm not sure if that answer relates to c or d best?

Answer 1

agree with d.

example: lim a->0 [sin(0+a)-sin(0)]/a=d(sin(x))/dx when x=0,,,,sin'(0)=cos(0)=1

the derivative limit is always of 0/0,,,,lim x->0 of g(x)-g(0)=0 always if g is continuous.

so answer c. is not correct,,, in example sin(0)=0

recall d(sin(x))/dx=lim a->0 [sin(x+a)-sin(x)]/a

=cos(x) because there is a removable discontinuity in the 0/0 form of the limit

[sin(x+a)-sin(x)]/a=[sinx(x)cos(a)cos(x)sin(a)-sin(x)]/a

when a->0, the right hand side becomes cos(x)sin(a)/a

and it can be shownm sin(a)/a approaches 1 when a->0

Answer 2

The limit is the formal definition of first derivative of function g(x) computed in x = 0; so (d) is correct. (e) is then incorrect.

No definite conclusion can be drawn for (a) or (b), since g(x) can be the smoothest function or the weirdest function. A for (c), any h(x) = g(x) + K, K constant, would still satisfy that relationship, and so the limit of g(x) as x approaches to 0 is undefined.