How to solve the derivative of (1/(x+3))+2x using limit as h approaches 0 definition?

Question

lim as h->0 of

 

(1/(x+3))+2x 

 

derivative definition help?

Answer 1

The definition of f'(x) you're referring to is the limit as h approaches 0 of [f(x+h) - f(x)]/h.

 

In this case we would take the limit of

 

[1/(x+h+3) + 2(x+h) - (1/(x+3) + 2x)]/h = [1/(x+h+3) + 2h - 1/(x+3)]/h.

 

As it stands the expression takes on the indeterminate form 0/0 after direct substitution of 0 for h.  To change the form I would simplify the complex fraction by multiplying each term by the common denominator (x+3)(x+h+3).

 

For example, to simplify [2/3 + 1 - 4/5]/2 we could multiply all four terms by 15 to get [10 + 15 - 12]/30.

 

After that focus on simplifying the numerator.  Each term in the numerator should contain a factor of h.  Hence, the numerator and denominator will have h as a common factor which can then be cancelled.  Since the resulting quotient will be continuous at 0 the limit can be found by direct substitution.

Answer 2

Is there a typo in this question? the variable here is x but you're stating that the limit is h approaching 0? 

 

In the case where it is x approaching 0, basically you just have to substitute the value of 0 for x in the expression.

 

(1/(x+3))+2x = (1/(0+3))+2(0) = (1/3) + 0 = 1/3

 

Hence, the answer would be 1/3.