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

Question

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

Scarlett Allison · Accepted Answer

The limit is #1#.

#Lim_(x -> 0)(3x)/(tan3x)# =#Lim_(x -> 0)(3x)/((sin3x)/(cos3x))# =#Lim_(x -> 0)(3xcos3x)/(sin3x)# =#Lim_(x -> 0)(3x)/(sin3x).cos3x# =#Lim_(x -> 0)color(red)((3x)/(sin3x)).cos3x# =#Lim_(x -> 0)cos3x# =#Lim_(x -> 0)cos(3*0)# #= Cos(0) = 1# Recall that: #Lim_(x -> 0)color(red)((3x)/(sin3x))=1# and #Lim_(x -> 0)color(red)((sin3x)/(3x))=1#

Luke Alvarez · Answer

undefined 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 $1$.

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