How to evaluate limx→05−5cos⁡(2x)+sin⁡(4x)x without using LHospitals rule? I thought the best way was to separate it in two limits: limx→0(5−5cos⁡(2x)x)+limx→0(sin⁡(4x)x) Considering that limx→0(sin⁡xx)=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 cos2x−sin2x, but it seems to be useless. So, to summarize everything, my problem is how to evaluate this limit without using LHospitals rule: limx→0(5−5cos⁡(2x)x)

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How to evaluate limx→05−5cos⁡(2x)+sin⁡(4x)x without using LHospitals rule?  I thought the best way was to separate it in two limits:  limx→0(5−5cos⁡(2x)x)+limx→0(sin⁡(4x)x)  Considering that limx→0(sin⁡xx)=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 cos2x−sin2x, but it seems to be useless.   So, to summarize everything, my problem is how to evaluate this limit without using LHospitals rule:  limx→0(5−5cos⁡(2x)x)

Brynn Ortiz · Accepted Answer

Hint: Use these two facts and a little bit of algebra:  limx→01−cos⁡(x)x=0  limx→0sin⁡(x)x=1  Note that  limx→010×1−cos⁡(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 1−cos⁡(x)=x22 when x is small and this proves the first relation.

Nevaeh Jensen · Accepted Answer

Hint:Using  cos⁡(2x)=cos2(x)−sin2(x)=1−sin2(x)−sin2(x)=1−2sin2(x)  we get  5−5cos⁡(2x)x=10sin⁡(x)⋅sin⁡(x)x

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

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