Use Theorem 3.10 to evaluate the following limits.
lim x...

Question

Use Theorem 3.10 to evaluate the following limits.

lim x🠂0 sin ax / sin bx, where a and b are constants with b ≠ 0.

Accepted Answer

Welcome back, everyone. Find the value of the limit. Limit as X approaches 0 of sin of 7 x divided by sin of 3 X. We're given 4 answer choices A says 7 divided by 3, B 3 divided by 7, C 1, and D 0. So for this problem, we want to recall that the limit as X approaches 0 of sine. Of x divided by X is equal to 1, and we're going to use this as a theorem that is going to help us solve for the given limit. What we can do is first we'll distribute the limit because we have a rational function, so we can take the limit off the top one, so that's limit as X approaches 0 off sign off. 7 of X and then divide that by the limit of the bottom function, which is limit as x approaches 0 of sin of 3 of X. If we use the direct substitution, well, immediately we're going to get 0 divided by 0, which is an indeterminate form. Instead, what we want to do is divide the top function by 7 of X. The reason why we're choosing this is because if we consider the theorem, we have to notice that sin of x divided by X simply means that we're taking sign of an angle and dividing by the same angle, right? So we're taking 7 of X because this is sin of 7 of X and this is our angle 7 of X. But if we're dividing by 7 of X, this means that we also have to multiply by 7 of X to make sure that the original fraction is unchanged, right? And now if x approaches 07 of X approaches 0 as well, right? And we're going to apply the same steps for the bottom one, so we have to choose 3 of X. We're dividing by 3 of X and multiplying by 3 of X. And now from here what we can do is simply notice that if we combine limits in a specific way using the properties of limits such that we take limits as X approaches 0 of 7 of X divided by 3 X. So 7 multiplied by X divided by 3. X we can notice that those Xs will cancel each other out and we will have limit as x approaches 0 of 7/3, which is independent of X, and gives us a constant 7/3 and then. We can distribute the limit as follows. Well, we're going to take limit as x approaches 0 of. Sign 7 X divided by 7 X. This gives us 1 by definition, and then at the bottom, we are going to have. Limit as X approaches 0 of sign. 3 X divided by 3X. This is also one based on the theorem, right? We can say that the top limit is equal to 1. And the bottom one is also equal to one. So we used a good manipulation of the properties of limits to get what we wanted, right? So we can say that this is equal to 7/3 because we have the first limit of a constant 7/3 multiplied by 1 divided by 1. So the correct answer to this problem corresponds to the answer choice A. Thank you for watching.

Use Theorem 3.10 to evaluate the following limits.lim x🠂0 sin ax / sin bx, where a and b are constants with b ≠ 0.

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Use Theorem 3.10 to evaluate the following limits.
lim x🠂0 sin ax / sin bx, where a and b are constants with b ≠ 0.