Limit Evaluation: Solve Lim (x->0) (-e^(2x)/x + 1/x + Sin(2019x)/x)
Hey guys! Let's dive into a fascinating problem today: evaluating the limit of a complex expression as x approaches 0. We're going to break down the expression, understand the behavior of each term, and then piece it all together to find the final limit. Buckle up, because this is going to be a fun ride through the world of calculus!
Understanding the Problem
So, the limit we need to tackle is:
lim (x->0) (-e^(2x)/x + 1/x + sin(2019x)/x)
At first glance, this might seem a bit intimidating. We've got exponential functions, fractions, and trigonometric functions all mixed up. But don't worry, we'll take it one step at a time. To evaluate this limit effectively, we need to understand how each term behaves as x approaches 0. Specifically, we need to look for indeterminate forms like 0/0 or ∞/∞, which might require us to use techniques like L'Hôpital's Rule or series expansions.
The first thing we should do is try to plug in x = 0 directly into the expression. This gives us:
-e^(20)/0 + 1/0 + sin(20190)/0 = -1/0 + 1/0 + 0/0
We can see that we have indeterminate forms (specifically, something involving division by zero), so we can't just directly substitute. This is our cue to dig a little deeper and use some limit evaluation techniques.
Let's rewrite the expression to combine some terms and see if it simplifies things. We can combine the first two terms since they have a common denominator:
lim (x->0) [(-e^(2x) + 1)/x + sin(2019x)/x]
This form might be a bit easier to work with. Now, let’s look at each part separately.
Analyzing the Components
1. The Term (-e^(2x) + 1)/x
Let’s first focus on the term (-e^(2x) + 1)/x. As x approaches 0, both the numerator and the denominator approach 0, giving us a 0/0 indeterminate form. This is a classic situation where L'Hôpital's Rule can be applied. L'Hôpital's Rule states that if we have a limit of the form lim (x->c) [f(x)/g(x)] where both f(x) and g(x) approach 0 or ±∞, then:
lim (x->c) [f(x)/g(x)] = lim (x->c) [f'(x)/g'(x)]
provided the limit on the right exists. So, we need to find the derivatives of the numerator and the denominator.
The derivative of the numerator, -e^(2x) + 1, with respect to x is:
d/dx (-e^(2x) + 1) = -2e^(2x)
The derivative of the denominator, x, with respect to x is:
d/dx (x) = 1
Now, we can apply L'Hôpital's Rule:
lim (x->0) (-e^(2x) + 1)/x = lim (x->0) (-2e^(2x))/1
Now, we can directly substitute x = 0:
-2e^(2*0)/1 = -2e^0 = -2
So, the limit of the first term as x approaches 0 is -2.
2. The Term sin(2019x)/x
Next, let's tackle the term sin(2019x)/x. This is another classic limit that you might recognize. We know that lim (x->0) sin(x)/x = 1. We can use this knowledge to evaluate our limit. We'll use a substitution to make it look more familiar.
Let u = 2019x. As x approaches 0, u also approaches 0. So, we can rewrite the limit as:
lim (x->0) sin(2019x)/x = lim (u->0) sin(u)/(u/2019)
We can rewrite this as:
lim (u->0) 2019 * sin(u)/u
Now, we can use the known limit lim (u->0) sin(u)/u = 1:
2019 * lim (u->0) sin(u)/u = 2019 * 1 = 2019
So, the limit of the second term as x approaches 0 is 2019.
Combining the Results
Now that we've found the limits of both components, we can combine them to find the limit of the entire expression. We found:
lim (x->0) (-e^(2x) + 1)/x = -2
and
lim (x->0) sin(2019x)/x = 2019
So, the original limit is:
lim (x->0) (-e^(2x)/x + 1/x + sin(2019x)/x) = lim (x->0) [(-e^(2x) + 1)/x + sin(2019x)/x]
= -2 + 2019
= 2017
Final Answer
Therefore, the limit of the expression as x approaches 0 is 2017. Woohoo! We did it!
Key Takeaways
- Indeterminate Forms: When evaluating limits, always check for indeterminate forms like 0/0 or ∞/∞. These forms indicate that you need to use techniques like L'Hôpital's Rule or series expansions.
- L'Hôpital's Rule: This rule is a powerful tool for evaluating limits of indeterminate forms. Remember to differentiate the numerator and the denominator separately.
- Known Limits: Recognizing and using known limits (like lim (x->0) sin(x)/x = 1) can simplify complex problems.
- Substitution: Sometimes, a clever substitution can transform a limit into a more manageable form.
- Breaking Down Problems: Complex problems often become easier when you break them down into smaller, more manageable parts.
Alternative Methods
While we used L'Hôpital's Rule and a known limit to solve this problem, there are other approaches we could have taken. For instance, we could have used Taylor series expansions to approximate the functions near x = 0. Let’s briefly look at how that would work.
Taylor Series Expansion
The Taylor series expansion of e^(2x) around x = 0 is:
e^(2x) = 1 + 2x + (2x)^2/2! + (2x)^3/3! + ...
So, -e^(2x) + 1 = -2x - 2x^2 - (4x^3)/3 - ...
Thus, (-e^(2x) + 1)/x = -2 - 2x - (4x^2)/3 - ...
As x approaches 0, this expression approaches -2, which matches our previous result.
The Taylor series expansion of sin(2019x) around x = 0 is:
sin(2019x) = 2019x - (2019x)^3/3! + (2019x)^5/5! - ...
So, sin(2019x)/x = 2019 - (2019^3 * x^2)/3! + ...
As x approaches 0, this expression approaches 2019, which also matches our previous result.
Using Taylor series expansions can be a powerful alternative to L'Hôpital's Rule, especially when dealing with more complex functions. It’s always good to have multiple tools in your calculus toolkit!
Practice Makes Perfect
Evaluating limits is a fundamental skill in calculus, and like any skill, it improves with practice. Try tackling similar problems and experimenting with different techniques. The more you practice, the more comfortable and confident you'll become.
So, keep practicing, keep exploring, and keep pushing your mathematical boundaries. You've got this!
Conclusion
Evaluating the limit lim (x->0) (-e^(2x)/x + 1/x + sin(2019x)/x) was a challenging but rewarding exercise. By breaking down the problem into smaller parts, applying L'Hôpital's Rule, and using known limits, we successfully found the limit to be 2017. We also explored an alternative method using Taylor series expansions, reinforcing the idea that there are often multiple paths to the same solution in calculus. I hope this explanation was helpful and insightful. Keep up the great work, and happy calculating!