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

Question

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

lim_{x \to 0} \frac{\sin x}{x} = 1

I am having hard time evaluating this limit:

lim_{x \to 0} \frac{\cos 3 x - 1}{\cos 2 x - 1}

I know, that this fact can help

lim_{x \to 0} \frac{\sin x}{x} = 1

But how? We can transform the first expression to

lim_{x \to 0} \frac{\cos 3 x - 1}{- 2 \sin^{2} x}

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

$lim_{x \to 0} \frac{\cos 3 x - 1}{\cos 2 x - 1} = \frac{9}{4} lim_{x \to 0} \frac{- \frac{1 - \cos 3 x}{9 x^{2}}}{- \frac{1 - \cos 2 x}{4 x^{2}}} = \frac{9}{4}$

This is helpful 0

sirpsta3u

Answered 2022-01-16 Author has 42 answers

Hint:
It is standard (from the limit of $\frac{\sin x}{x}$ ) that
$lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2}, s o 1 - \cos x \sim_{0} \frac{x^{2}}{2}$ ,
and you may proceed by substitution.