0
$\begingroup$

I am setting up the problem to find the limit as delta x goes to zero, using the definition in the correct format. No matter what I've tried, I am stuck without finding the right cancellations to leave me with $-4/x^{1.5}$

Would you help me get unstuck?

I have tried everything I can think of and I am missing some little algebra technique. Any help would be much appreciated.

Added Note: A word to other calculus and precalculus beginners - don't forget how to tie your shoes just because you're working with new ideas like Limits and derivatives. Algebra still works, but you will encounter new opportunities for ingenuity not usually seen in most precalculus or algebra classes.

1 Answers 1

1

$$\frac1h\left(\frac8{\sqrt{x+h}}-\frac8{\sqrt{x}}\right)=\frac{-8}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}.$$ Hint: $$\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$$

  • 2
    Thanks for trying to help me. Where I am stuck is when I work with the left hand expression, I multiplied through by the product of the denominators to create a common denominator. What I am left with still gives me 0/0. I do not understand how you got your right hand side, nor what to do with it to arrive at my known answer. Do you see where I got stuck? A gentle nudge would be appreciated.2011-10-10
  • 0
    Once you write the difference as a single fraction, there sould be a term of the form $\sqrt{a}-\sqrt{b}$ in the numerator. Nudge: use the hint on this term.2011-10-10
  • 0
    What is the function you are trying to find the limit for?2011-10-10
  • 0
    Emmad, it is 8 / (x^1/2)2011-10-10
  • 0
    @FreeT, no it is not. You are trying to find the limit of the ratio $(f(x+h)-f(x))/h$ when $h\to0$.2011-10-10
  • 1
    Yes, I understand that Didier. I think I should not have taken a shortcut when answering to Emmad. your hint above is a very good hint Didier. Is this term a regular feature when multiplying by a conjugate?2011-10-10
  • 0
    Please tell me if the nudge worked for you or if I should add more detailed explanations.2011-10-10
  • 0
    I don't understand the hint part. I think after the second step of your solution, substituting 'h' for 0 in the fraction yields the required derivative. So, can you clarify the use of the hint you gave later?2018-08-01
  • 0
    @MrReality 1. The hint is designed to help reach the identity above thus, if you can prove the identity above, then it is not useful to you. 2. Substituting $h=0$ in the fraction is not the way to compute the derivative. Rather, consider the limit when $h\to0$. In the present case, it happens that both procedures yield the same result but nevertheless, one is rigorous and the other is not.2018-08-01