Use l’Hôpital’s Rule to find the limits in Exercises 85–108.

Question

99. lim(x→0) (2^sin(x) - 1)/(e^x - 1)

Accepted Answer

Welcome back, everyone. Evaluate the limit using Lapito's rule. Limit as X approaches 0 of 3 the power of sine x minus 1 divided by E power of 2X minus 1. A says LN of 3 divided by 3, BLN of 3 divided by 2, CLN of 3, and D 2 LN of 3. For this problem, let's rewrite the limit as x approaches 0 of 3 to the power of sine x minus 1 divided by e to the power of 2 x minus 1. And let's remember that the first step is to substitute X equals 0 for every X. So we get 3 to the power of sine of 0 minus 1 divided by E to the power of 2 multiplied by 0 minus 1. Now, sign of 0 is 0, with the power of 0 is 1, and 1 minus 1 is 0. We got 0 in the numerator. And the denominator eats the power of 0 is 1 and 1 minus 1 is also 0. 0 divided by 0 is a suitable form for Lapito's rule to be applied. And we can now proceed with the rule. According to the rule, we want to differentiate the numerator. And we want to find the derivative. Of that denominator. Let's find each. We're going to get limit as x approaches 0. Now, 3 to the power of sine X is a composite function. First of all, we are differentiating an exponential function with respect to sin X. To get 3 to the power of sine X. LN of 3. And according to the chain rule, we're multiplying by the derivative of sine x, right? The derivative of sine x is cosine, so we can just write cosine x. Now The derivative of -1 is 0 because it's a constant, so we're not including it. And in the denominator, we have another composite function, it's the power of 2 X. We are differentiating with respect to 2 X, which is the inner function to get eats the power of 2 X, multiplied by the derivative of the inner function. So again, we're applying the chain rule. And the derivative of 2 X is simply 2, right? So we can write 2. E to the power of 2 X. The derivative of -1 is 0. Well then, so from here we can again perform the right substitution which gives us 3. So the power of sin of 0, which is 0. Multiplied by a lot of 3, multiplied by cosine of 0. Cosine of 0 is 1, right? And we're dividing by 2 to the power of 2 multiplied by 0 gives us 0. Now let's simplify.wes the power of 0 is 1. 1 multiplied by 1 is 1, so it basically has no effect. Each the power of 0 is 1, so we simply get a line of 3 divided by 2, right, which is finite. At one of 3 divided by 2, and that's our final answer for this problem. Which corresponds to the answer choice B. Thank you for watching.

Use l’Hôpital’s Rule to find the limits in Exercises 85–108.99. lim(x→0) (2^sin(x) - 1)/(e^x - 1)

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Use l’Hôpital’s Rule to find the limits in Exercises 85–108.
99. lim(x→0) (2^sin(x) - 1)/(e^x - 1)