Limits And Integrals: Step-by-Step Solutions

Hey guys! Today, we're diving deep into the fascinating world of limits and integrals. We'll break down how to evaluate some tricky limits and integrals, making sure you understand every step along the way. So, buckle up and let's get started!

Limit Evaluation: A Step-by-Step Approach

Evaluating Limits at Infinity

Our first challenge is to evaluate the limit as x approaches infinity for the rational function:

lim (x→∞) (3x^2 - 2x + 1) / (5 - 8x^2)

When dealing with limits at infinity for rational functions, the key is to focus on the highest powers of x in the numerator and denominator. In this case, that's x².

To tackle this, we'll divide both the numerator and the denominator by x². This might sound a bit scary, but trust me, it simplifies things beautifully. Here’s how it looks:

lim (x→∞) [(3x^2 - 2x + 1) / x^2] / [(5 - 8x^2) / x^2]

Now, let's distribute the division:

lim (x→∞) [3 - (2/x) + (1/x^2)] / [(5/x^2) - 8]

As x approaches infinity, the terms 2/x, 1/x², and 5/x² all approach zero. Think about it: if you divide a constant by an increasingly large number, the result gets closer and closer to zero. This leaves us with:

lim (x→∞) [3 - 0 + 0] / [0 - 8] = 3 / -8 = -3/8

So, the limit as x approaches infinity for the given function is -3/8. See? Not so intimidating after all!

This method works because as x grows infinitely large, the terms with lower powers of x become insignificant compared to the highest power. By dividing through by the highest power, we effectively isolate the dominant terms and make the limit much easier to evaluate.

Key Takeaway: When evaluating limits of rational functions at infinity, divide both the numerator and denominator by the highest power of x to simplify the expression.

Evaluating Limits with Radicals

Next up, we've got a limit that involves a radical:

lim (x→4) (x - 4) / (√x - 2)

If we try to directly substitute x = 4, we end up with 0/0, which is an indeterminate form. This tells us we need to do some algebraic manipulation to find the limit. The trick here is to rationalize the denominator. This means we'll multiply both the numerator and the denominator by the conjugate of the denominator.

The conjugate of √x - 2 is √x + 2. So, let's multiply:

lim (x→4) [(x - 4) / (√x - 2)] * [(√x + 2) / (√x + 2)]

Multiplying the numerators and denominators, we get:

lim (x→4) [(x - 4)(√x + 2)] / [(√x - 2)(√x + 2)]

The denominator simplifies nicely using the difference of squares formula (a - b)(a + b) = a² - b²:

lim (x→4) [(x - 4)(√x + 2)] / [x - 4]

Now, we can cancel the (x - 4) terms:

lim (x→4) (√x + 2)

Finally, we can substitute x = 4:

√4 + 2 = 2 + 2 = 4

So, the limit as x approaches 4 for the given function is 4. Rationalizing the denominator allowed us to eliminate the indeterminate form and find a definite limit.

Key Takeaway: When dealing with limits involving radicals, rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator.

Integral Evaluation: Mastering the U-Substitution

Now, let's switch gears and tackle an integral:

∫[-2, 2] x * e^(x^2) dx

This integral looks a bit intimidating at first, but we can use a clever technique called u-substitution to make it much more manageable. U-substitution is like the chain rule in reverse, and it's super handy for integrals where you see a function and its derivative (or a multiple of its derivative) within the integral.

In this case, we notice that the derivative of x² is 2x, and we have an x in the integral. This is a good sign that u-substitution will work! Let's set:

u = x^2

Then, we find the differential du:

du = 2x dx

We can rearrange this to solve for x dx:

(1/2) du = x dx

Now, we need to change the limits of integration from x values to u values. When x = -2:

u = (-2)^2 = 4

And when x = 2:

u = (2)^2 = 4

Notice that both limits of integration are now 4. This is a crucial observation! Our integral now becomes:

∫[4, 4] (1/2) e^u du

Wait a minute! The limits of integration are the same. What does that mean? Well, it means the integral is zero!

Key Insight: The integral of any function over an interval of zero length is always zero. This is because the integral represents the area under the curve, and there's no area if the interval has no width.

So, without even needing to find the antiderivative of e^u, we know that:

∫[-2, 2] x * e^(x^2) dx = 0

This result makes sense if we think about the function f(x) = x e^(x2). This function is an odd function because f(-x) = -f(x). The integral of an odd function over a symmetric interval (like -2 to 2) is always zero. The areas on either side of the y-axis cancel each other out.

Key Takeaway: When evaluating definite integrals, always check if the function is odd or even and if the interval is symmetric. This can save you a lot of work!

Conclusion: Mastering Limits and Integrals

So there you have it! We've successfully evaluated a limit at infinity, a limit with radicals, and a tricky integral using u-substitution. Remember, the key to mastering limits and integrals is practice, practice, practice! Keep working at it, and you'll become a pro in no time. Keep exploring those mathematical concepts, guys!