Using limθ→0 sin θ / θ = 1

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Question

Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limx→0 (x² − x + sin x) / 2x

Accepted Answer

Welcome back, everyone. Find the limit of the function F of T equals 3T2 + 2 T minus sine of T divided by 5 T as T approaches 0. We're given 4 answer choices AS 2/5, B 1/5, C 3/5, and D 5. So let's begin by evaluating the limit. Limit as T approaches 0 of 3 T squad. Plus 2 t minus sine of T divided by 5 T. We can begin with the rank substitution which would give us 3 multiplied by 0 plus 2 multiplied by 0 minus sign of 0. Our numerator simply becomes 0, and our denominator as C approaches 0 is 0 as well. 5 multiplied by 0 is 0. So we get an indeterminate form if we assume that our function is continuous at T equals 0, it is not, right? So this means that we have to somehow evaluate the limit using a different way, right? And because we have a fraction which essentially involves T2 T and sin of T, those are different functions. One of them is a trigonometric function. We're going to simply apply the properties of limits. We're going to separate. Our T2 and T from sine. So first of all, we get limit as T approaches 0 of 3 T2 + 2T divided by 5T, right? We can simply apply the properties of fractions, which is going to give us 2 fractions, and then we're distributing the limit for each fraction using the properties of limits, right? So, first of all, we're applying the limit for 3 T2 plus 2T divided by 5T, and then we are subtracting the limit as T approaches 0 of sin of T divided by 5T. Soon we will understand why this is useful. For the first limit, we can perform division, right? So we get limit as T approaches 0 of 3 T 2 divided by 5T. This gives us 3/5 T. T 2 divided by T gives us T, and then we have a ratio between the coefficients, and then 2T divided by 5T gives us 2 5s. For the second limit, we can notice that we have sin of T divided by T. So what we want to do is simply factor out the constant, which is 1/5 multiplied by limit as T approaches 0 of sine of T divided by T. Now, let's focus on the second limit. By definition, if T approaches 0, the limit of sin of T divided by T is equal to 1, right? So we can say that this limit is equal to 1. We're going to use a basic definition, and then for the first limit, we're going to perform direct substitution. So this gives us 3/5 multiplied by 0 + 25. And then we are subtracting 1/5 multiplied by 1, right, because our last limit is equal to 1. And now we can perform the calculation, so we end up with 2/5s. Minus 1/5, and this is equal to 1/5. Which essentially corresponds to the answer choice B. Thank you for watching.

Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limx→0 (x² − x + sin x) / 2x

Key Concepts

Limit of a Function

Sine Function and Its Limit

L'Hôpital's Rule

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