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. limx→0(1√1+x−1x)
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:
limx→0(1√1+x−1x)=limx→0(1√1+x−√1+x√1+xx)
limx→0(1√1+x−1x)=limx→0(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:
limx→0(1√1+x−1x)=limx→0(x+1)1/2[1−(x+1)1/2]x(x+1)
limx→0(1√1+x−1x)=limx→01−(x+1)1/2x(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. limΔx→0sin(π6+Δx)−12Δx
I am not really sure what to do here.
+9490 1.
=limx→0(1√1+x−1x) =limx→0(1√1+x−√1+x√1+xx) =limx→0(1−√1+x√1+xx) =limx→0(1−√1+x√1+x÷x) =limx→0(1−√1+x√1+x⋅1x) =limx→0(1−√1+xx√1+x)
When you get to this point, instead of multiplying the numerator and denominator by √1+x ,
try multiplying the numerator and denominator by 1+√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.
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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?
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)=sinAcosB+cosAsinB.
limΔx→0sin(π6+Δx)−12Δx=limΔx→0sin(π6)cosΔx+cos(π6)sinΔx−12ΔxlimΔx→0sin(π6+Δx)−12Δx=limΔx→012cosΔx+√32sinΔx−12ΔxlimΔx→0sin(π6+Δx)−12Δx=limΔx→0cosΔx+√3sinΔx−12Δx
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.
+26399 1.
limx→0(1√1+x−1x)
limx→0(1√1+x−1x)=limx→0((1+x)−12−1x)|L'Hospital's rule=limx→0(d ((1+x)−12−1)dxd (x)dx)=limx→0(−12(1+x)−12−11)|x→0=−12
+130508 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
