I feel like I am understanding the concept of limits. I have not had many problems with solving limits, but there are a few problems that are giving me more grief than I anticipated. I really do not want someone to spoonfeed me the answer here; please only steer me in the right direction. Thanks!


1. \(\lim_{x \to 0}\left(\frac{\frac{1}{\sqrt{1+x}}-1}{x}\right)\)


I have tried a few things to no avail. I tried combining this complex fraction into a simpler one. Here's what I have tried:


\(\lim_{x \to 0}\left(\frac{\frac{1}{\sqrt{1+x}}-1}{x}\right)=\lim_{x \to 0}\left(\frac{\frac{1}{\sqrt{1+x}}-\frac{\sqrt{1+x}}{\sqrt{1+x}}}{x}\right)\)

\(\lim_{x \to 0}\left(\frac{\frac{1}{\sqrt{1+x}}-1}{x}\right)=\lim_{x \to 0}\frac{(x+1)^{1/2}-(x+1)}{x(x+1)}\)

I now tried factoring out the largest common divider from the numerator and see what happens:

\(\lim_{x \to 0}\left(\frac{\frac{1}{\sqrt{1+x}}-1}{x}\right)=\lim_{x \to 0}\frac{(x+1)^{1/2}\left[1-(x+1)^{1/2}\right]}{x(x+1)}\)

\(\lim_{x \to 0}\left(\frac{\frac{1}{\sqrt{1+x}}-1}{x}\right)=\lim_{x \to 0}\frac{1-(x+1)^{1/2}}{x(x+1)^{1/2}}\)


This does not appear to have served me any good. I have not made any progress. I cannot substitute 0 in for the argument of the limit as it still outputs indeterminate. There must be some other method I am neglecting, no?


2. \(\lim_{\Delta x \to 0}\frac{\sin\left(\frac{\pi}{6}+\Delta x\right)-\frac{1}{2}}{\Delta x}\)


I am not really sure what to do here. 

 Sep 2, 2019



\(\phantom{=\quad}\lim\limits_{x\to0}\left(\dfrac{\frac{1}{\sqrt{1+x}}-1}{x}\right)\\~\\ {=\quad}\lim\limits_{x\to0}\left(\dfrac{\frac{1}{\sqrt{1+x}}-\frac{\sqrt{1+x}}{\sqrt{1+x}}}{x}\right)\\~\\ {=\quad}\lim\limits_{x\to0}\left(\dfrac{\frac{1-\sqrt{1+x}}{\sqrt{1+x}}}{x}\right)\\~\\ {=\quad}\lim\limits_{x\to0}\left({\frac{1-\sqrt{1+x}}{\sqrt{1+x}}}\div{x}\right)\\~\\~\\ {=\quad}\lim\limits_{x\to0}\left({\frac{1-\sqrt{1+x}}{\sqrt{1+x}}}\cdot\frac{1}{x}\right)\\~\\~\\ {=\quad}\lim\limits_{x\to0}\left({\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}}\right)\)


When you get to this point, instead of multiplying the numerator and denominator by   \(\sqrt{1+x}\) ,


try multiplying the numerator and denominator by   \(1+\sqrt{1+x}\)   (the conjugate of the numerator)


After you do that, you should be able to simplify it to a point where you can substitute  0  in for  x  and get a defined value.




Hope this helps for your first question!

As for your second question, have you learned about or are you allowed to use L'Hopital's rule?

 Sep 2, 2019

1. I am sort of surprised that the concept of the conjugate did not spring into my mind. Thanks for that small tip. 


2. No, I have not learned L'Hopital's rule. There is a hint near the footnote of the page for this problem; it says to think about trigonometric identities. It seems to me that the sum-difference identity is the proper one to use. Specifically, \(\sin(A+B)=\sin A\cos B+\cos A\sin B\).


\(\lim_{\Delta x \to 0}\frac{\textcolor{red}{\sin\left(\frac{\pi}{6}+\Delta x\right)}-\frac{1}{2}}{\Delta x}=\lim_{\Delta x \to 0}\frac{\textcolor{red}{\sin\left(\frac{\pi}{6}\right)\cos \Delta x+\cos\left(\frac{\pi}{6}\right)\sin \Delta x}-\frac{1}{2}}{\Delta x}\\ \lim_{\Delta x \to 0}\frac{\sin\left(\frac{\pi}{6}+\Delta x\right)-\frac{1}{2}}{\Delta x}=\lim_{\Delta x \to 0}\frac{\frac{1}{2}\cos \Delta x+\frac{\sqrt{3}}{2}\sin \Delta x-\frac{1}{2}}{\Delta x}\\ \lim_{\Delta x \to 0}\frac{\sin\left(\frac{\pi}{6}+\Delta x\right)-\frac{1}{2}}{\Delta x}=\lim_{\Delta x \to 0}\frac{\cos \Delta x+\sqrt{3}\sin \Delta x-1}{2\Delta x}\\ \)


I still do not see a way to simplify this limit. Although, I may be closer than I was before. The indeterminate form still exists. 

Guest Sep 3, 2019


\(\lim \limits_{x \to 0}\left(\dfrac{\dfrac{1}{\sqrt{1+x}}-1}{x}\right) \)


\(\begin{array}{|rcll|} \hline && \mathbf{\lim \limits_{x \to 0}\left(\dfrac{\dfrac{1}{\sqrt{1+x}}-1}{x}\right) } \\\\ &=&\lim \limits_{x \to 0}\left(\dfrac{ \left(1+x \right)^{-\frac{1}{2}} -1}{x}\right) \quad | \quad \text{L'Hospital's rule} \\\\ &=&\lim \limits_{x \to 0} \left( \dfrac{ \dfrac{ d\ \left( \left(1+x \right)^{-\frac{1}{2}} -1 \right) } { dx } } { \dfrac{d\ (x)}{dx} } \right) \\\\ &=&\lim \limits_{x \to 0} \left( \dfrac{ -\frac{1}{2} \left(1+x \right)^{-\frac{1}{2}-1} } {1} \right) \quad | \quad x\to 0 \\\\ &=& \mathbf{-\dfrac{1}{2}} \\ \hline \end{array}\)




 Sep 3, 2019

You are going in the right direction.....we can split your answer  up into these  functions


         cosΔx - 1                             sinΔx 

(1/2)  _______    +    (√ 3/2)    _________ 

            Δx                                       Δx




We have two identities to remember  :


As   x →  0,     cos Δx  -  1

                       __________     approaches 0......



So   the limit of the first function  just simplifies to  0


Also    as x   → 0 ,     sin Δx

                                  ______   approaches  1......



So the limit of the second function just simplifies  to (√ 3/2) 





lim               sin ( pi/6  +  Δx)  - 1/2           

 x→ 0         __________________   =   (√ 3/2) 




See the graph here : https://www.desmos.com/calculator/mnpxjiqitj



cool cool cool

 Sep 3, 2019
edited by CPhill  Sep 3, 2019

Thank you, Cphill!


When limits are involved, shuffling terms can be quite powerful. This strategy has let me finish the rest of the packet as it is a common theme throughout the limit packet I am completing. 

Guest Sep 5, 2019

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