How to evaluate limit without L'Hospital's rule, but using \lim_{x \to 0} \frac{\sin x}{x}=1 N

Danelle Albright 2022-01-14 Answered
How to evaluate limit without LHospitals rule, but using limx0sinxx=1
I am having hard time evaluating this limit:
limx0cos3x1cos2x1
I know, that this fact can help
limx0sinxx=1
But how? We can transform the first expression to
limx0cos3x12sin2x
but what should I do next? I cannot use LHopitals rule for solving this.
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Answers (2)

Lindsey Gamble
Answered 2022-01-15 Author has 38 answers

limx0cos3x1cos2x1=94limx01cos3x9x21cos2x4x2=94

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sirpsta3u
Answered 2022-01-16 Author has 42 answers

Hint:
It is standard (from the limit of sinxx) that
limx01cosxx2=12,so 1cosx0x22,
and you may proceed by substitution.

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