Understanding Absolute Value Limits

Hey guys, let's dive into a super common math problem that pops up in calculus: limits involving absolute value functions. Specifically, we're going to break down the limits of $\frac{|x-1|}{x-1}$ as $x$ approaches 4 from both the left and the right, and then figure out if the overall limit exists at that point. This might seem a bit tricky at first, but once you get the hang of how absolute values work with limits, it's a piece of cake!

Approaching from the Left: $\lim _{x \rightarrow 4^{-}} \frac{|x-1|}{x-1}$

Alright, first up, we're looking at the limit as $x$ approaches 4 from the left. This means we're considering values of $x$ that are just slightly less than 4. Think numbers like 3.9, 3.99, 3.999, and so on. When we're dealing with absolute values, the key is to determine whether the expression inside the absolute value is positive or negative in the region we're considering. In this case, the expression is $(x-1)$.

Since $x$ is approaching 4 from the left, $x$ is always going to be less than 4 (but greater than 1, which is what matters here). If $x$ is less than 4, then $(x-1)$ will be less than $(4-1)$, which is 3. More importantly, since $x$ is approaching 4, $x$ will definitely be greater than 1. This means that $(x-1)$ will always be a positive number in this scenario. For instance, if $x = 3.9$, then $(x-1) = 2.9$, which is positive. If $x = 3.999$, then $(x-1) = 2.999$, still positive!

Because $(x-1)$ is positive when $x$ approaches 4 from the left, the absolute value $|x-1|$ simplifies to just $(x-1)$. So, for values of $x$ close to 4 but less than 4, our expression $\frac{|x-1|}{x-1}$ becomes $\frac{x-1}{x-1}$.

Now, here's the cool part: for any value of $x$ where $x \neq 1$ (which is true for $x$ approaching 4), the expression $\frac{x-1}{x-1}$ simplifies to just 1. Since we're looking at the limit as $x$ approaches 4, $x$ is definitely not 1. Therefore, $\frac{x-1}{x-1} = 1$ for all the $x$ values we're interested in.

So, the limit becomes $\lim _{x \rightarrow 4^{-}} 1$. The limit of a constant is just that constant. That means:

$\lim _{x \rightarrow 4^{-}} \frac{|x-1|}{x-1} = 1$

Pretty straightforward, right? The absolute value didn't even change anything because the term inside it was positive.

Approaching from the Right: $\lim _{x \rightarrow 4^{+}} \frac{|x-1|}{x-1}$

Next up, let's tackle the limit as $x$ approaches 4 from the right. This means we're considering values of $x$ that are just slightly greater than 4. Think numbers like 4.1, 4.01, 4.001, and so on. Again, we need to check the sign of the expression inside the absolute value, which is $(x-1)$.

Since $x$ is approaching 4 from the right, $x$ is always going to be greater than 4. If $x$ is greater than 4, then $(x-1)$ will be greater than $(4-1)$, which is 3. Any number greater than 4, when you subtract 1 from it, will result in a number greater than 3. For example, if $x = 4.1$, then $(x-1) = 3.1$, which is positive. If $x = 4.001$, then $(x-1) = 3.001$, still positive!

Because $(x-1)$ is positive when $x$ approaches 4 from the right, the absolute value $|x-1|$ again simplifies to just $(x-1)$. So, for values of $x$ close to 4 but greater than 4, our expression $\frac{|x-1|}{x-1}$ also becomes $\frac{x-1}{x-1}$.

Just like in the previous case, since we're looking at the limit as $x$ approaches 4, $x$ is not equal to 1. Thus, $\frac{x-1}{x-1}$ simplifies to 1 for all the $x$ values we're considering.

Therefore, the limit becomes $\lim _{x \rightarrow 4^{+}} 1$. And as we know, the limit of a constant is the constant itself.

So, the limit is:

$\lim _{x \rightarrow 4^{+}} \frac{|x-1|}{x-1} = 1$

Wow, it looks like the limit from the right is also 1! This is because, in the vicinity of $x=4$ (whether approaching from the left or right), the expression $(x-1)$ is always positive, making the absolute value redundant.

Does the Overall Limit Exist? $\lim _{x \rightarrow 4} \frac{|x-1|}{x-1}$

Now for the million-dollar question: does the overall limit, $\lim _{x \rightarrow 4} \frac{|x-1|}{x-1}$, exist? For a limit to exist at a certain point, the function must approach the same value as $x$ approaches that point from both the left side and the right side. In other words, the left-hand limit must be equal to the right-hand limit.

Let's recall what we found:

• The limit from the left: $\lim _{x \rightarrow 4^{-}} \frac{|x-1|}{x-1} = 1$
• The limit from the right: $\lim _{x \rightarrow 4^{+}} \frac{|x-1|}{x-1} = 1$

Since the limit from the left (1) is exactly equal to the limit from the right (1), we can confidently say that the overall limit exists at $x=4$. And not only does it exist, but its value is also 1.

So, the answer to whether the limit exists is Yes, and the limit is 1.

Key Takeaway:

When evaluating limits involving absolute values, always pay close attention to the sign of the expression inside the absolute value bar as $x$ approaches the limit point. This determines whether $|expression|$ simplifies to $expression$ or $-expression$. In the case of $\frac{|x-1|}{x-1}$, the expression $(x-1)$ is positive for all $x$ values around 4 (both slightly less and slightly more). This means $|x-1| = x-1$ in this specific scenario, simplifying the entire fraction to 1 for $x \neq 1$. This is why both the left-hand and right-hand limits, and thus the overall limit, are equal to 1. Keep practicing these, and you'll master them in no time! You guys got this!