# How do you find the limit of #(3(1-cosx))/x# as #x->0#?

You can use the de l'Hospital's rule to find this limit. See explanation.

Generally speaking the rule says that instead of calculating original limit

you can calculate:

(i.e. instead of the limit of quotient of 2 functions you calculate the limit of quotient of their first derivatives)

and those limits will either both be equal or neither of them will exist.

Note: If the limit of quotient of first derivatives is still undefined you can repeat this procedure (calculate the limit of quotient of 2nd derivatives).

Here we have:

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A fundamental trigonometric limit is

A second important trigonometric limit is

So for this question,

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To find the limit of (3(1-cosx))/x as x approaches 0, we can use L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately with respect to x.

Differentiating the numerator, we get: d/dx (3(1-cosx)) = 3(0 - (-sinx)) = 3sinx.

Differentiating the denominator, we get: d/dx (x) = 1.

Now, we can evaluate the limit of the differentiated function as x approaches 0.

lim(x->0) (3sinx)/1 = 3sin(0)/1 = 0/1 = 0.

Therefore, the limit of (3(1-cosx))/x as x approaches 0 is 0.

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