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Functions continuous at specific x-values

Determine the continuity of two functions, ln(x-3) and e^(x-3), at x=3. Explore the concept of continuity, highlighting that common functions are continuous within their domain. Discover that ln(x-3) is not continuous at x=3, while e^(x-3) is continuous for all real numbers, including x=3.

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  • hopper happy style avatar for user Jacqueline
    Im sure this is a silly question and Im missing the obvious, but why did you say g(3)=ln(0)? How did you know to put zero? Are you saying that nothing raised to a power equals three therefore its zero? Thank you!
    (11 votes)
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  • leaf green style avatar for user Ananya Kundu
    what is meant by the natural logarithm?
    (11 votes)
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  • blobby green style avatar for user Yames Yamb
    On the practice for this section, it asks us to know if a function is continuous for all real numbers.. one of the options is

    cubic root of (x+1)

    And it is at f(x) at -2... how can you take a cubic root of -1? isn't that an imaginary value? and thus not real?
    (7 votes)
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    • duskpin sapling style avatar for user Vu
      Just as square root is the inverse of squaring, cube root is the inverse of cubing.
      Cube root of -1 is -1, because (-1)^3 = -1 * -1 * -1 = -1

      You can't have a square root of a negative number, this would result in imaginary number. This is true and extends to all even roots, i.e. square root, 4th root, 6th root, so on and so on. But imaginary number only applies to even roots. You can have cube root or any odd roots of a negative number. Cube root of -1 is one example. Cube root of -8 is -2, because -2 * -2 * -2 = -8.

      Hope that helps.
      (15 votes)
  • hopper cool style avatar for user Jake
    What if a specific x-value is defined but the point still not continuous? Obvoisly, you could tell that the point is not continous by graphing the function, but is there a way to tell just by looking at the equation? Also, is there an example of an equation where a point is defined but not continuous?
    (10 votes)
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  • duskpin ultimate style avatar for user Sadat Tameem
    In the Practise, I found this hint for a question that doesn't make sense to me. It says: f(x)=tan(x)=sin(x)/cos(x) is defined for all numbers except when cos(x) is 0. It makes sense-you can't divide by zero but then it says: so it's domain is all x-values such that x does not equal *(2k+1)Xpi/2* where k is an integer.

    How do u get *(2k+1)Xpi/2*? Please help!
    (5 votes)
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    • piceratops ultimate style avatar for user Fai
      So, you understand the basic idea. f(x)=tan(x) is not defined when cos(x) =0, because tan(x) is sin(x)/cos(x).

      Let's look at where cos(x) =0. You'll need to draw the graph of cos(x) for yourself. You'll see that in a 2pi interval, there are two points where cos(x) = 0. Those points are pi/2 and 3pi/2.

      Now, if you were to extend the graph further, you'd see that 5pi/2, 7pi/2 are also zeros.

      So, let's write down a sequence which defines where cos(x) =0:
      pi/2, 3pi/2, 5pi/2, 7pi/2...

      Notice anything about the sequence? They're odd numerators times pi with denominator 2.

      So, the formula then becomes:
      (2k+1)Pi/2 where k is any integer
      (11 votes)
  • orange juice squid orange style avatar for user Marvin Humphrey
    Is it fair to say that while the function f(x) = √x is continuous on the interval [0,∞), it is not continuous at the point x = 0 because the two-sided limit doesn't exist?
    (3 votes)
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  • starky sapling style avatar for user Nightmare252
    At x = 3 he says it is undefined and draws an asymptote how does the function cross a vertical asymptote? I thought functions can only cross horizontal and slant asymptotes and can never cross vertical ones
    (3 votes)
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  • blobby green style avatar for user sreedhar.iyer
    At 4.09, the graph is shown extending into negative values. Is this correct? As x takes on higher negative values, y values asymptotically approach zero, isn't it? In fact there is no way a negative value can be found for function f(x)
    (2 votes)
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  • blobby green style avatar for user hs21h006
    How will we identify whether a function is defined or not?
    (1 vote)
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    • male robot donald style avatar for user Venkata
      A function itself cannot be defined/undefined. However, it can be defined/undefined *at a point* (the terminology is important).

      To check for whether it is defined or not, simply plug that point into the function. For example, if I have $f(x) = \frac{1}{x}$ and x = 0, I get $f(0) = \frac{1}{0}$. Clearly, $\frac{1}{0}$ isn't defined. So, the function isn't defined at x = 0, which is clearly shown by the graph as it asymptotes towards infinity as x gets closer to 0.
      (4 votes)
  • aqualine ultimate style avatar for user Reza
    Why ln(0) doesn't work but e^0 works?
    (1 vote)
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    • starky ultimate style avatar for user KLaudano
      If e^x=y, then ln(y)=x. e^0=1 and ln(1)=0, so there are no contradictions.

      If ln(0) was defined, this would imply that there exists some value, n, such that e^n=0. However, e^x > 0 for all values of x. So, the value n does not exist and the natural log of 0 is undefined.
      (2 votes)

Video transcript

- [Voiceover] Which of the following functions are continuous at x equals three? Well, as we said in the previous video, in the previous example, in order to be continuous at a point, you at least have to be defined at that point. We saw our definition of continuity, f is continuous at a, if and only if, the limit of f as x approaches a is equal to f of a. So, over here, in this case, we could say that a function is continuous at x equals three, so f is continuous at x equals three, if and only if the limit as x approaches three of f of x, is equal to f of three. Now let's look at this first function right over here. Natural log of x minus three. Well, try to evaluate it, and it's not an f now, it's g, try to evaluate g of three. G of three, let me write it here, g of three is equal to the natural log of zero. Three minus three. This is not defined. You can't raise e to any power to get to zero. You can try to go to, you could say, negative infinity, but that's not, this is not defined. And so, if this isn't even defined at x equals three, there's no way that it's going to be continuous at x equals three, so we could rule this one out. Now f of x is equal to e to the x minus three. Well this is just a shifted over version of e to the x. This is defined for all real numbers, and as we saw in the previous example, it's reasonable to say it's continuous for all real numbers, and you could even do this little test here. The limit of e to the x minus three as x approaches three, well that is going to be, that is going to be e to the three minus three, or e to the zero, or one. And so f is the only one that is continuous. And once again, it's good to think about what's going on here visually, if you like. Both of these are, you could think of them, this is a shifted over version of ln of x, this is a shifted over version of e to the x, and so if we like, we could draw ourselves some axes, so that's our y-axis, this is our x-axis, and actually, let me draw some points here. So that's one, that is one, that is two, three, two, and three, and let's see, I said these are shifted over versions, so actually, this is maybe not the best way to draw it, so let me draw it, this is one, two, three, four, five, and six. And on this axis, I won't make 'em on the same scale, let's say this is one, two, three. I'm gonna draw one, two, three, I'm gonna draw a dotted line right over here. So g of x, ln of x minus three is gonna look something like this. If you put three in it, it's not defined, if you put four in it, ln of four, well, that's gonna, sorry, ln of four minus one, so that's gonna be ln of four minus three, is actually let me just draw a table here, I know I'm confusing you. So, if I say x and I say g of x, so at three, you're undefined. At four, this is ln of one, ln of one, which is equal to zero, so it's right over there. So g of x is gonna look something like, something like that. And so you can see at three, you have this discontinuity there, it's not even defined to the left of three. Now f of x is a little bit more straightforward. If you have, so e to the three is going to be, sorry, f of three is going to be e to the three minus three, or e to the zero, so it's going to be one, so it's gonna look something like this, it's gonna look something, something like, like that. There's no jumps, there's no gaps, it is going to be continuous, and frankly, all real numbers so for sure it's going to be continuous at three.