Use series to evaluate the limit.
limit as x approach 0
[1-cos(5x)]/[1+5x-e^(5x)]
To evaluate the limit as x approaches 0 of the given expression, [1 - cos(5x)] / [1 + 5x - e^(5x)], we can use a Taylor series expansion.
1. Begin by expanding the numerator, 1 - cos(5x), into its Maclaurin series. Recall that the Maclaurin series for cos(x) is given by:
cos(x) = 1 - (x^2 / 2!) + (x^4 / 4!) - (x^6 / 6!) + ...
In our case, we substitute 5x for x in the series:
1 - cos(5x) = 1 - [1 - (5x)^2 / 2! + (5x)^4 / 4! - (5x)^6 / 6! + ...]
Simplifying this expression gives:
1 - cos(5x) = 5x^2 / 2! - 5x^4 / 4! + 5x^6 / 6! - ...
2. Next, we expand the denominator, 1 + 5x - e^(5x), into its Maclaurin series. Since e^(x) has a well-known Maclaurin series, we can use that. The Maclaurin series for e^(x) is given by:
e^(x) = 1 + x + (x^2 / 2!) + (x^3 / 3!) + (x^4 / 4!) + ...
In our case, we substitute 5x for x in the series:
e^(5x) = 1 + 5x + (5x)^2 / 2! + (5x)^3 / 3! + (5x)^4 / 4! + ...
Simplifying this expression gives:
e^(5x) = 1 + 5x + 25x^2 / 2! + 125x^3 / 3! + 625x^4 / 4! + ...
3. Now, we substitute the expanded expressions into the original limit expression:
[1 - cos(5x)] / [1 + 5x - e^(5x)] = [5x^2 / 2! - 5x^4 / 4! + 5x^6 / 6! - ...] / [1 + 5x - (1 + 5x + 25x^2 / 2! + 125x^3 / 3! + 625x^4 / 4! + ...)]
4. We can simplify the expression by canceling out common terms:
[5x^2 / 2! - 5x^4 / 4! + 5x^6 / 6! - ...] / [1 + 5x - (1 + 5x + 25x^2 / 2! + 125x^3 / 3! + 625x^4 / 4! + ...)]
= [5x^2 / 2! - 5x^4 / 4! + 5x^6 / 6! - ...] / [-25x^2 / 2! - 125x^3 / 3! - 625x^4 / 4! + ...]
5. Now, we can simplify further by dividing each term by x^2:
[5x^0 / 2! - 5x^2 / 4! + 5x^4 / 6! - ...] / [-25x^0 / 2! - 125x^1 / 3! - 625x^2 / 4! + ...]
6. Taking the limit as x approaches 0:
lim(x->0) [5x^0 / 2! - 5x^2 / 4! + 5x^4 / 6! - ...] / [-25x^0 / 2! - 125x^1 / 3! - 625x^2 / 4! + ...]
= [5(0)^0 / 2! - 5(0)^2 / 4! + 5(0)^4 / 6! - ...] / [-25(0)^0 / 2! - 125(0)^1 / 3! - 625(0)^2 / 4! + ...]
= 0 / 0 (indeterminate form)
Using series expansion, we have obtained an indeterminate form. To evaluate the limit further, additional techniques might be required, such as applying L'Hopital's Rule or simplifying the expression using algebraic manipulations.
To evaluate the limit as x approaches 0 for the given expression, we can use a Taylor series expansion.
Let's start by expanding the numerator, 1 - cos(5x), using the Maclaurin series for cosine:
cos(x) = 1 - (x^2/2!) + (x^4/4!) - (x^6/6!) + ...
Replacing x with 5x, we get:
cos(5x) = 1 - ((5x)^2/2!) + ((5x)^4/4!) - ((5x)^6/6!) + ...
= 1 - (25x^2/2!) + (625x^4/4!) - (15625x^6/6!) + ...
Now let's expand the denominator, 1 + 5x - e^(5x), using the Maclaurin series for exponential function:
e^x = 1 + x + (x^2/2!) + (x^3/3!) + ...
Replacing x with 5x, we get:
e^(5x) = 1 + (5x) + ((5x)^2/2!) + ((5x)^3/3!) + ...
= 1 + 5x + (25x^2/2!) + (125x^3/3!) + ...
Now substitute both expansions back into the original expression:
[1 - cos(5x)] / [1 + 5x - e^(5x)]
= [1 - (25x^2/2!) + (625x^4/4!) - (15625x^6/6!)] / [1 + 5x + (25x^2/2!) + (125x^3/3!) + ...]
Taking the limit as x approaches 0 means we can neglect the terms with higher powers of x.
After simplifying, we have:
lim(x->0) [1 - cos(5x)] / [1 + 5x - e^(5x)]
= [1 - (25x^2/2!)] / [1 + 5x]
Now, we can substitute x=0 into the simplified expression:
[1 - (25(0)^2/2!)] / [1 + 5(0)]
= [1 - 0] / [1 + 0]
= 1 / 1
= 1
Therefore, the limit as x approaches 0 for the given expression is 1.