Is This Function Continuous At A=1? Let's Check!

Hey math whizzes! Today, we're diving deep into the world of calculus and tackling a super common question: how do we determine if a function is continuous at a specific point? We've got a neat little function here, and a trusty continuity checklist to guide us. So, buckle up, grab your notebooks, and let's get this done!

Understanding Continuity: The Big Picture

Alright guys, before we jump into our specific problem, let's quickly chat about what continuity even means in math. Think of a continuous function like a line you can draw without lifting your pen. There are no sudden jumps, breaks, or holes. It just flows smoothly. In calculus, we formalize this idea with a three-part checklist. If a function passes all three tests at a particular point, we say it's continuous at that point. If it fails even one test, then bam! It's discontinuous. It's like a bouncer at a club – you gotta meet all the criteria to get in. And understanding this concept is absolutely fundamental for so many areas of math, from graphing and understanding limits to more advanced topics like integration and differential equations. So, really nailing down this continuity idea is a game-changer, trust me.

Now, let's get to our specific function and see if it makes the cut. Our function is defined piecewise, meaning it has different rules for different values of x. This is super common and often where we find points of interest (and potential discontinuities!). Here's what we're working with:

$$f(x)=\left\{\begin{array}{ll} \frac{x^2-1}{x-1} & \text { if } x \neq 1 \\ 10 & \text { if } x=1 \end{array}\right.; a=1$$

See how it's defined one way for x not equal to 1, and another way when x is exactly 1? This is exactly the kind of setup that requires us to put our continuity hat on and go through the checklist systematically. We're going to check the continuity at a=1, so that's our target point. Let's break down the checklist and apply it step-by-step. It’s like solving a puzzle, and each condition is a piece we need to fit together.

The Continuity Checklist: Three Essential Conditions

So, what exactly is this magic checklist I keep talking about? Well, for a function f(x) to be continuous at a point x = a, it MUST satisfy these three conditions:

1. The function must be defined at a: This means that f(a) must exist and have a specific, numerical value. You can't have a hole or an undefined point right where you're trying to check for continuity. If f(a) is undefined (like dividing by zero, or taking the square root of a negative number in the real number system), then the function is automatically discontinuous at a.

2. The limit of the function as x approaches a must exist: This means that as x gets really, really close to a (from both sides!), the function's output, f(x), must approach a single, specific number. We write this as lim_(x->a) f(x) = L, where L is a finite number. If the limit doesn't exist – maybe the function approaches different values from the left and right, or it shoots off to infinity – then it's discontinuous.

3. The limit must equal the function's value: This is the final check, and it ties everything together. The value the function approaches as x gets close to a (the limit, L) must be exactly the same as the actual value of the function at a (which is f(a)). So, we need lim_(x->a) f(x) = f(a). If the limit exists and the function is defined, but these two values don't match, then we have a 'removable discontinuity', which is still a type of discontinuity.

Remember these three points, guys. They are your bread and butter for determining continuity. Missing any one of them means the function is discontinuous. It’s crucial to understand that each condition is necessary, but not sufficient on its own. You need all three to line up perfectly. Let's put this checklist to work on our example function.

Step 1: Is the Function Defined at a=1? 🤔

Okay, first things first. We need to check if our function f(x) is defined at our point of interest, which is a=1. Looking back at the definition of f(x):

$$f(x)=\left\{\begin{array}{ll} \frac{x^2-1}{x-1} & \text { if } x \neq 1 \\ 10 & \text { if } x=1 \end{array}\right.$$

See that second line? It explicitly tells us what f(x) is when x is exactly equal to 1. It says f(1) = 10. Since we have a specific, numerical value (10) for f(1), the first condition is met! Yes, the function is defined at a=1, and its value is 10. This is great news and means we can move on to the next step. If this step had failed, we could have stopped right there and declared the function discontinuous. But we passed, so let's keep going!

Step 2: Does the Limit Exist as x Approaches 1? 🧐

Now for the second condition: does the limit of f(x) as x approaches 1 exist? This means we need to investigate what value f(x) gets close to as x gets close to 1, but not actually equal to 1. For this, we use the first part of our function's definition, where x ≠ 1:

$$\lim_{x \to 1} \frac{x^2-1}{x-1}$$

If we try to just plug in x=1 here, we get (1^2 - 1) / (1 - 1), which is 0/0. Uh oh! That's an indeterminate form, meaning we can't determine the limit directly by substitution. This is a classic sign that we need to simplify the expression. Remember your algebra, folks! The numerator, x^2 - 1, is a difference of squares. It can be factored as (x-1)(x+1).

So, we can rewrite the limit expression like this:

$$\lim_{x \to 1} \frac{(x-1)(x+1)}{x-1}$$

Since x is approaching 1 but is not equal to 1, x-1 is not zero. This means we can cancel out the (x-1) terms in the numerator and denominator!

$$\lim_{x \to 1} (x+1)$$

Now, this simplified expression is much easier to evaluate. We can just substitute x=1 into (x+1):

$$1 + 1 = 2$$

So, the limit of the function as x approaches 1 is 2. Since we found a specific, finite number (2) that the function approaches, yes, the limit exists, and its value is 2. We've passed another crucial test!

Step 3: Does the Limit Equal the Function's Value? 🤔

We're on the home stretch, guys! The final condition is the tie-breaker: does the limit we just found equal the actual value of the function at a=1?

From Step 1, we know that f(1) = 10.

From Step 2, we found that lim_(x->1) f(x) = 2.

Now we compare these two values: Is 2 equal to 10?

No, 2 is definitely not equal to 10.

This means that our function fails the third condition for continuity. Even though the function is defined at x=1 and the limit exists as x approaches 1, the limit does not equal the function's value at that point. We call this a removable discontinuity. It's like there's a little 'oops' right at x=1. If f(1) had been defined as 2, then the function would have been continuous there. But as it stands, it's not.

Conclusion: Continuous or Not?

So, let's recap our journey through the continuity checklist for our function f(x) at a=1:

1. Is f(1) defined? Yes, f(1) = 10.
2. Does lim_(x->1) f(x) exist? Yes, lim_(x->1) f(x) = 2.
3. Does lim_(x->1) f(x) = f(1)? No, 2 ≠ 10.

Since the function fails the third condition, we can definitively say that the function f(x) is NOT continuous at a=1.

It's important to remember that discontinuities aren't necessarily