# Lim 3x/tan3x x →0 How to solve it ? I think the answer will be 1 or -1 who can solve it ?

The limit is

Recall that:

and

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To solve the limit (\lim_{x \to 0} \frac{3x}{\tan 3x}), we can use L'Hôpital's Rule, which states that if the limit of the quotient of two functions is of the form (\frac{0}{0}) or (\frac{\pm \infty}{\pm \infty}), then the limit of their derivatives is the same as the limit of the original functions.

First, let's find the derivatives of (3x) and (\tan 3x):

[\frac{d}{dx}(3x) = 3] [\frac{d}{dx}(\tan 3x) = 3\sec^2(3x)]

Now, apply L'Hôpital's Rule:

[\lim_{x \to 0} \frac{3x}{\tan 3x} = \lim_{x \to 0} \frac{3}{3\sec^2(3x)}]

Now, as (x) approaches (0), (\sec^2(3x)) approaches (\sec^2(0) = 1). Therefore,

[\lim_{x \to 0} \frac{3}{3\sec^2(3x)} = \frac{3}{3 \cdot 1} = 1]

So, the limit is indeed (1).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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