Limits of Average Rates of Change

Question

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form limh→0 (f(x+h) − f(x)) / h occur frequently in calculus. In Exercises 57–62, evaluate this limit for the given value of x and function f.

f(x) = x², x = -2

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Determine the limit of the expression. F of X plus H minus F of X divided by H as H approaches 0 for the function F of X is equal to 5 x 2, and evaluate it at X equals 3. Fantastic. So it appears for this particular problem we're asked to determine the limit of this expression that is provided to us, and then we're asked to Evaluate it at X equals 3, provided that we're told that the function F of X is equal to 5x the power of 2. So now that we know what we're ultimately trying to solve for, let's read off our multiple choice answers to see what our final answer might be. A is 24, B is 12, C is 30, and D is 27. OK. So first off, let us write the limit that we're trying to determine. Let's write down our limit that we're trying to find. So the limit as H approaches 0 of. F of X plus H minus F of X all divided by H. And once again we're trying to find it where X is equal to 3. So let us first solve for F of X plus H. So let us note that F of X plus H, as we should recall, will be equal to 5 multiplied by X plus H. And once again, we're plugging in our expression. So we're trying to solve for F of X plus H using our original function F of X. So we need to take 5 multiplied by X + H2. Which is going to be equal to 5 multiplied by X to the 2 + 2 X H plus H2, which is gonna be all equal to, let me simplify, 5 X2 + 10 XH plus 5 H2. Fantastic. So, now we need to substitute in our values for F of X and F of X plus H into our limit expression and simplify. So we need to take the limit as H approaches 0 of. F, so right now we're taking our original function up above and we're plugging in our known variables. So now we know that F of X plus H is going to be equal to 5 X2 + 10. XH. Plus 5 H2 minus 5 X2. All divided by H, and this is going to be equal to the limit as H approaches 0 up when we do a little bit of simplifying. 10 XH plus 5 H squared divided by H, and this will be all equal to the limit as H approaches 0 up, H multiplied by 10 X plus 5 H divided by H. Which will be equal to the limit as H approaches 0, 10 X plus 5 H. Fantastic. And then this will be all equal to. Square brackets, 10 X plus 5. Multiplied by 0 because we're plugging in a, when we do direct substitution plugging in a value of 0 in for H, that will mean that we yield an answer of 10 X. And since we are told, now we're told that since X equals 3, that will mean that if we take 10 multiplied by 3, our final answer has to be simply 30. So therefore, the limit is 30, and this corresponds to multiple choice answer letter C, which once again is 30. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Limits of Average Rates of Change

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form limh→0 (f(x+h) − f(x)) / h occur frequently in calculus. In Exercises 57–62, evaluate this limit for the given value of x and function f.

f(x) = x², x = -2

Key Concepts

Limits

Derivatives

Secant and Tangent Lines

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