146k views
3 votes
Lim x→0 (4x − sin(4x)/(4x − tan(4x)))

User Kamesh
by
7.9k points

1 Answer

5 votes

Answer:

To find the limit of the given expression as x approaches 0, you can use L'Hôpital's Rule, which states that if the limit of the ratio of two functions as x approaches a limit point is of the form 0/0 or ∞/∞, then the limit can be found by taking the derivative of the numerator and denominator separately and then evaluating the limit again.

Let's apply L'Hôpital's Rule to your expression:

Lim(x→0) [(4x - sin(4x)) / (4x - tan(4x))]

First, find the derivatives of the numerator and denominator:

Numerator:

d/dx (4x - sin(4x)) = 4 - 4cos(4x)

Denominator:

d/dx (4x - tan(4x)) = 4 - 4sec^2(4x)

Now, we have:

Lim(x→0) [(4 - 4cos(4x)) / (4 - 4sec^2(4x))]

Now, you can evaluate the limit:

Lim(x→0) (4 - 4cos(4x)) / (4 - 4sec^2(4x))

Now, plug in x = 0:

(4 - 4cos(0)) / (4 - 4sec^2(0))

Cos(0) = 1 and sec(0) = 1:

(4 - 4) / (4 - 4 * 1^2)

(0) / (0)

Now, we have an indeterminate form 0/0 again. So, we need to apply L'Hôpital's Rule one more time.

Take the derivatives of the numerator and denominator again:

Numerator:

d/dx (4 - 4cos(4x)) = 16sin(4x)

Denominator:

d/dx (4 - 4sec^2(4x)) = 16sec(4x)tan(4x)

Now, evaluate the limit again:

Lim(x→0) (16sin(4x)) / (16sec(4x)tan(4x))

Now, plug in x = 0:

(16sin(0)) / (16sec(0)tan(0))

Sin(0) = 0, sec(0) = 1, and tan(0) = 0:

(16 * 0) / (16 * 1 * 0)

This simplifies to 0/0, which is still an indeterminate form.

Let's apply L'Hôpital's Rule one more time:

Numerator:

d/dx (16sin(4x)) = 64cos(4x)

Denominator:

d/dx (16sec(4x)tan(4x)) = 64sec(4x)tan^2(4x)

Now, evaluate the limit again:

Lim(x→0) (64cos(4x)) / (64sec(4x)tan^2(4x))

Now, plug in x = 0:

(64cos(0)) / (64sec(0)tan^2(0))

Cos(0) = 1, sec(0) = 1, and tan(0) = 0:

(64 * 1) / (64 * 1 * 0)

This simplifies to 64/0, which is an indeterminate form of ∞/0.

So, the limit of the given expression as x approaches 0 is ∞/0, which means the limit is infinity.

Explanation:

User Pejuko
by
7.9k points