Question

1) lim (x~~0) (sin3x-sin2x)/(sin5x-sin4x) 2) lim (x~~0) ((2^x)-1)/((3^x)-1)

Answer 1

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Answer 2

Let's find the limits for both of these expressions:

1. lim (x→0) (sin(3x) - sin(2x)) / (sin(5x) - sin(4x))

To solve this limit, we can use L'Hôpital's Rule. L'Hôpital's Rule states that if you have a limit of the form 0/0 or ∞/∞, you can take the derivative of the numerator and denominator separately and then re-evaluate the limit.

First, let's differentiate the numerator and denominator:

Numerator: d/dx [sin(3x) - sin(2x)] = 3cos(3x) - 2cos(2x)

Denominator: d/dx [sin(5x) - sin(4x)] = 5cos(5x) - 4cos(4x)

Now, we can re-evaluate the limit as x approaches 0:

lim (x→0) [(3cos(3x) - 2cos(2x)) / (5cos(5x) - 4cos(4x))]

Plugging in x = 0:

(3cos(0) - 2cos(0)) / (5cos(0) - 4cos(0))

Simplifying further:

(3 - 2) / (5 - 4)

1/1 = 1

So, the limit of the given expression as x approaches 0 is 1.

2. lim (x→0) ((2^x) - 1) / ((3^x) - 1)

This limit can be evaluated directly without the need for L'Hôpital's Rule. When x approaches 0, you can plug in the value of x:

((2^0) - 1) / ((3^0) - 1) = (1 - 1) / (1 - 1) = 0/0

Notice that you have an indeterminate form 0/0. In this case, you can try to simplify the expression further. We can rewrite the expression as follows:

((2^x) - 1) / ((3^x) - 1) = ((2^x) - 1) / ((3^x) - 1) * (1/1)

Now, let's use the limit property that lim (x→0) 1 = 1:

lim (x→0) ((2^x) - 1) / ((3^x) - 1) * lim (x→0) 1

The limit of a constant is the constant itself, so:

1 * 1 = 1

So, the limit of the given expression as x approaches 0 is 1.