How to evaluate \lim _{x\to 0}\frac{5-5\cos(2x)+\sin(4x)}{x} without using L'Hospital's rule?

mairie0708zl

mairie0708zl

Answered question

2022-01-28

How to evaluate limx055cos(2x)+sin(4x)x without using LHospitals rule?
I thought the best way was to separate it in two limits:
limx0(55cos(2x)x)+limx0(sin(4x)x)
Considering that limx0(sinxx)=1 we can easily know that the second limit is 4. I also know the result of the original limit is 4, so the result of the first limit in the second line needs to be 0 (0+4=4).
The issue is that I can not figure out how to remove the x from the denominator so I can avoid the indeterminate form. I already tried replacing cos(2x) with cos2xsin2x, but it seems to be useless.
So, to summarize everything, my problem is how to evaluate this limit without using LHospitals rule:
limx0(55cos(2x)x)

Answer & Explanation

Brynn Ortiz

Brynn Ortiz

Beginner2022-01-29Added 12 answers

Hint: Use these two facts and a little bit of algebra:
limx01cos(x)x=0
limx0sin(x)x=1
Note that
limx010×1cos(2x)2x=0
The most straightforward and rigorous way to see why the first relation holds is to use the Taylor series for cosine. A non-rigorous way of seeing the first relation holds, if we accept the second relation, besides using the double-angle formula and similar trigonometric relations as pointed out by others, is this non-rigorous but more or less intuitive argument:
When x is small, we know that:
sin(x)x
cos(x)=sin(x)x22+C
If x=0, then cos(x)=1, hence C=1
This shows that 1cos(x)=x22 when x is small and this proves the first relation.
Nevaeh Jensen

Nevaeh Jensen

Beginner2022-01-30Added 14 answers

Hint:Using
cos(2x)=cos2(x)sin2(x)=1sin2(x)sin2(x)=12sin2(x)
we get
55cos(2x)x=10sin(x)sin(x)x

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