Calculating Limits: A Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of limits in calculus. We'll break down three different limit problems step-by-step, so you can see exactly how to solve them. Grab your calculators and let's get started!

a. Limit of (5 - y)^(4/3) as y approaches -3

Let's kick things off with our first limit problem:

$$\lim_{y \to -3}(5 - y)^{4/3}$$

In this case, we're asked to find the limit of the function $(5 - y)^{4/3}$ as $y$ approaches $-3$. The great thing about this function is that it's continuous at $y = -3$. This means we can simply substitute $-3$ for $y$ to find the limit.

Here's how we do it:

1. Substitute $-3$ for $y$ in the function:

   $(5 - (-3))^{4/3} = (5 + 3)^{4/3} = 8^{4/3}$

2. Simplify:

   Now we need to evaluate $8^{4/3}$. Remember that $a^{m/n}$ is the same as $\sqrt[n]{a^m}$. So, $8^{4/3}$ is the same as $\sqrt[3]{8^4}$.

   • First, find the cube root of 8: $\sqrt[3]{8} = 2$
   • Then, raise that result to the power of 4: $2^4 = 16$

So, $8^{4/3} = 16$.

Therefore,

$$\lim_{y \to -3}(5 - y)^{4/3} = 16$$

In conclusion, for this limit, direct substitution works perfectly because the function is continuous at the point we're approaching. This makes the calculation straightforward and easy to understand. Remember, when faced with a limit, always check if direct substitution is a valid method before trying anything more complex!

b. Limit of (u^6 - 6u^3 + 9) / (u^3 + u^4 - 3 - 3u) as u approaches 3√3

Next up, we have a more challenging limit:

$$\lim_{u \to 3\sqrt{3}} \frac{u^6 - 6u^3 + 9}{u^4 + u^3 - 3u - 3}$$

This limit looks intimidating because if we directly substitute $u = 3\sqrt{3}$, we'll likely end up with an indeterminate form like $\frac{0}{0}$. So, we need to manipulate the expression to simplify it before evaluating the limit.

Here’s how we can approach it:

1. Factor the numerator and denominator:

   • Numerator: $u^6 - 6u^3 + 9$ can be seen as a quadratic in terms of $u^3$. Let $x = u^3$. Then the numerator becomes $x^2 - 6x + 9$, which factors to $(x - 3)^2$. Replacing $x$ with $u^3$, we get $(u^3 - 3)^2$.
   • Denominator: $u^4 + u^3 - 3u - 3$ can be factored by grouping. Group the first two terms and the last two terms: $u^3(u + 1) - 3(u + 1)$. Now, factor out the common factor of $(u + 1)$, giving us $(u + 1)(u^3 - 3)$.

   So, the expression becomes:

   $$\lim_{u \to 3\sqrt{3}} \frac{(u^3 - 3)^2}{(u + 1)(u^3 - 3)}$$

2. Simplify the expression:

   Notice that we have a common factor of $(u^3 - 3)$ in both the numerator and the denominator. We can cancel one of these factors out:

   $$\lim_{u \to 3\sqrt{3}} \frac{u^3 - 3}{u + 1}$$

3. Substitute $u = 3\sqrt{3}$:

   Now we can try substituting $u = 3\sqrt{3}$ into the simplified expression:

   $$\frac{(3\sqrt{3})^3 - 3}{3\sqrt{3} + 1}$$

4. Evaluate:

   • $(3\sqrt{3})^3 = 3^3 * (\sqrt{3})^3 = 27 * 3\sqrt{3} = 81\sqrt{3}$

   • So the expression becomes:

     $$\frac{81\sqrt{3} - 3}{3\sqrt{3} + 1}$$

5. Rationalize the denominator (optional, but often helpful):

   To rationalize the denominator, multiply both the numerator and denominator by the conjugate of the denominator, which is $3\sqrt{3} - 1$:

   $$\frac{(81\sqrt{3} - 3)(3\sqrt{3} - 1)}{(3\sqrt{3} + 1)(3\sqrt{3} - 1)}$$

   • Expanding the numerator:

     $(81\sqrt{3} - 3)(3\sqrt{3} - 1) = 81\sqrt{3} * 3\sqrt{3} - 81\sqrt{3} - 9\sqrt{3} + 3 = 81 * 3 * 3 - 90\sqrt{3} + 3 = 729 - 90\sqrt{3} + 3 = 732 - 90\sqrt{3}$

   • Expanding the denominator:

     $(3\sqrt{3} + 1)(3\sqrt{3} - 1) = (3\sqrt{3})^2 - 1^2 = 9 * 3 - 1 = 27 - 1 = 26$

   • So, the expression becomes:

     $$\frac{732 - 90\sqrt{3}}{26}$$

6. Simplify further:

   Divide both terms in the numerator by 2 (since both are even numbers):

   $$\frac{366 - 45\sqrt{3}}{13}$$

Therefore,

$$\lim_{u \to 3\sqrt{3}} \frac{u^6 - 6u^3 + 9}{u^4 + u^3 - 3u - 3} = \frac{366 - 45\sqrt{3}}{13}$$

In conclusion, this problem required a few more steps than the first one. Factoring, simplifying, and then rationalizing the denominator were all key to solving this limit. Remember, don't be afraid to manipulate the expression to make it easier to evaluate.

c. Limit of (3t + 7) / (t - 9) as t approaches ∞

Finally, let's tackle a limit as t approaches infinity:

$$\lim_{t \to \infty} \frac{3t + 7}{t - 9}$$

When dealing with limits at infinity, we often look at the highest powers of the variable in the numerator and denominator. In this case, both the numerator and denominator have a highest power of $t^1$ (just $t$).

Here's how we solve it:

1. Divide both the numerator and denominator by the highest power of $t$:

   In this case, the highest power is $t$. So, we divide both the numerator and denominator by $t$:

   $$\lim_{t \to \infty} \frac{\frac{3t}{t} + \frac{7}{t}}{\frac{t}{t} - \frac{9}{t}} = \lim_{t \to \infty} \frac{3 + \frac{7}{t}}{1 - \frac{9}{t}}$$

2. Evaluate the limit as $t$ approaches infinity:

   As $t$ approaches infinity, the terms $\frac{7}{t}$ and $\frac{9}{t}$ approach 0:

   $$\lim_{t \to \infty} \frac{3 + \frac{7}{t}}{1 - \frac{9}{t}} = \frac{3 + 0}{1 - 0} = \frac{3}{1} = 3$$

Therefore,

$$\lim_{t \to \infty} \frac{3t + 7}{t - 9} = 3$$

In conclusion, when dealing with limits at infinity, dividing by the highest power of the variable is a crucial step. This simplifies the expression and allows us to easily see what happens as the variable grows without bound. Remember, terms like $\frac{c}{t}$ (where $c$ is a constant) will always approach 0 as $t$ approaches infinity.

So there you have it! We've tackled three different types of limit problems, each requiring a slightly different approach. Whether it's direct substitution, factoring, or dividing by the highest power, understanding these techniques will help you conquer any limit that comes your way. Keep practicing, and you'll become a limit-solving pro in no time! Cheers, guys!