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

How to evaluate limit without LHospitals rule, but using $\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$
I am having hard time evaluating this limit:
$\underset{x\to 0}{lim}\frac{\mathrm{cos}3x-1}{\mathrm{cos}2x-1}$
I know, that this fact can help
$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$
But how? We can transform the first expression to
$\underset{x\to 0}{lim}\frac{\mathrm{cos}3x-1}{-2{\mathrm{sin}}^{2}x}$
but what should I do next? I cannot use LHopitals rule for solving this.
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$\underset{x\to 0}{lim}\frac{\mathrm{cos}3x-1}{\mathrm{cos}2x-1}=\frac{9}{4}\underset{x\to 0}{lim}\frac{-\frac{1-\mathrm{cos}3x}{9{x}^{2}}}{-\frac{1-\mathrm{cos}2x}{4{x}^{2}}}=\frac{9}{4}$

###### Not exactly what you’re looking for?
sirpsta3u

Hint:
It is standard (from the limit of $\frac{\mathrm{sin}x}{x}$) that
,
and you may proceed by substitution.