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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
 #1
avatar+8965 
+4

1.

 

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

 

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
 #3
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+1

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
 #2
avatar+25225 
+2

1.

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

 

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

 

 

laugh

 Sep 3, 2019
 #4
avatar+111387 
+2

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......

                                Δx 

 

So   the limit of the first function  just simplifies to  0

 

Also    as x   → 0 ,     sin Δx

                                  ______   approaches  1......

                                     Δx

 

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

 

 

So  

 

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

 x→ 0         __________________   =   (√ 3/2) 

                            Δx

 

 

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

 

 

cool cool cool

 Sep 3, 2019
edited by CPhill  Sep 3, 2019
 #5
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0

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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