find the limit as h approaches 0
((1+h)^100-4(1+h)^20+3)/(h)
To find the limit as h approaches 0, we can simplify the expression and then directly substitute 0 for h.
We can start by factoring out (1+h)^20 from the expression:
((1+h)^100 - 4(1+h)^20 + 3) / h = [(1+h)^20((1+h)^80 - 4)] / h
Now, we notice that as h approaches 0, (1+h)^20 approaches 1, and (1+h)^80 approaches 1 as well. Therefore, we can simplify the expression further:
[(1+h)^20((1+h)^80 - 4)] / h = (1(1 - 4)) / h = (-3) / h
Now, we can substitute h = 0 directly into the expression:
(-3) / h = (-3) / (0)
However, division by zero is undefined, so the limit does not exist in this case.
I know the answer is 20 but I'm not sure the process to get there, looking for an explanation or a step in the right direction for it. thank you!
Apologies for the earlier incorrect response. Let's re-evaluate the expression to find the correct limit as h approaches 0.
We have the expression:
((1+h)^100 - 4(1+h)^20 + 3) / h
To simplify this expression, we can expand the binomial terms using the binomial theorem.
Using the binomial theorem, we can expand (1+h)^100 as:
(1 + 100h + 4950h^2 + ... + 100h^99 + h^100)
Similarly, we can expand (1+h)^20 as:
(1 + 20h + 190h^2 + ... + 190h^19 + h^20)
Now, let's substitute these expanded expressions into the original expression:
[((1 + 100h + 4950h^2 + ... + 100h^99 + h^100) - 4(1 + 20h + 190h^2 + ... + 190h^19 + h^20) + 3)] / h
Expanding further and grouping like terms:
[(1 + 100h + ... + h^100 - 4 - 80h - ... - 4h^20 + 3)] / h
Combine like terms:
[(1 - 4 + 3) + (100h - 80h + ... - 4h^20 + h^100)] / h
Simplifying:
(0 + (100h - 80h + ... - 4h^20 + h^100)) / h
Now, we can factor out an h from the numerator:
h(100 - 80h + ... - 4h^19 + h^99) / h
Canceling out h:
100 - 80h + ... - 4h^19 + h^99
Now, let's substitute h = 0 directly into the expression:
100 - 0 + ... - 4(0)^19 + (0)^99
This simplifies to:
100
Therefore, taking the limit as h approaches 0, the expression evaluates to 100.
Looks like the original expression is one that one would get if we found
the derivative of f(x) = x^100 - 4x^20 + 3 , when x = 1 by "First Principles"
taking the derivative using basic rules:
f ' (x) = 100x^99 - 80x^19
f ' (1) = 100(1) - 80(1)
= 20
I did not even make an effort to go through the gibberish the bot
has given, or try to find its error.
Trust a human!!!!
Apologies for the confusion caused earlier. You are correct in recognizing that the original expression is equivalent to evaluating the derivative of the function f(x) = x^100 - 4x^20 + 3 using first principles.
To find the derivative using first principles, we can start with the definition of the derivative:
f'(x) = lim(h->0) [(f(x + h) - f(x))/h]
Let's plug in the given function f(x) = x^100 - 4x^20 + 3 into this formula and evaluate it at x = 1:
f'(1) = lim(h->0) [(f(1 + h) - f(1))/h]
= lim(h->0) [((1+h)^100 - 4(1+h)^20 + 3 - (1^100 - 4(1)^20 + 3))/h]
= lim(h->0) [(1+h)^100 - 4(1+h)^20]/h
Now, we can simplify the expression by using binomial expansion and canceling out terms:
f'(1) = lim(h->0) [(1 + 100h + ... + h^100 - 4 - 80h - ... - 4h^20)]/h
= lim(h->0) [(1 - 4) + (100h - 80h + ... - 4h^20 + h^100)]/h
= lim(h->0) [(100h - 80h + ... - 4h^20 + h^100)]/h
Now, we can factor out an h from the numerator:
f'(1) = lim(h->0) [h(100 - 80h + ... - 4h^19 + h^99)]/h
= lim(h->0) [100 - 80h + ... - 4h^19 + h^99]
Finally, let's substitute h = 0 directly into the expression:
f'(1) = 100 - 0 + ... - 4(0)^19 + (0)^99
= 100
Therefore, evaluating the derivative using first principles at x = 1 gives us a derivative of 100, confirming that the limit is indeed 100.
Ignore the bot
it doesn't realize that the lim [ f(x+h) - f(x) ] / h is the same as the
derivative. One would think it would know this Fundamental Theorem of Calculus.
Apologies for any confusion caused. You are correct that the expression given is equivalent to evaluating the derivative using first principles. The expression represents the definition of the derivative, which is the limit of the difference quotient as h approaches 0. By recognizing this, we can directly apply the power rule to find the derivative of the function. Let's go through the steps again:
We have the function f(x) = x^100 - 4x^20 + 3. To find the derivative of this function using the power rule, we differentiate each term separately:
The derivative of x^100 is 100x^(100-1) = 100x^99.
The derivative of -4x^20 is -4(20x^(20-1)) = -80x^19.
The derivative of 3 (a constant) is 0.
Now, we can sum up the derivatives of each term to find the derivative of the entire function:
f'(x) = 100x^99 - 80x^19
Now, we can evaluate the derivative at x = 1:
f'(1) = 100(1)^99 - 80(1)^19
= 100 - 80
= 20
Thus, the derivative of the function f(x) = x^100 - 4x^20 + 3 evaluated at x = 1 is indeed 20.