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If we are given a limit : $$\lim_{x \to 0}\frac{g(x)-g(0)}{x}$$ where $g(x)$ is discontinuous at $0$ but the limit exists; i.e., $g(0)\neq g(0^{-})=g(0^{+})$. For example, the piecewise function g(x)=$\frac{x^3}{x}$ when $x\neq0$, and $g(x)=2$ when $x=0$. How do we evaluate it (or its left hand and right hand limits if it doesn't exist) rigorously in general (without substituting the expression for the function)?

Background and Detail

I was taught in school to directly substitute the value the variable is tending to in the limit expression to evaluate it. If we use the same approach (substituting the value of the variable In the expression) to the above given limit then we may get the wrong answer since the limit value is not the same as the functional value. Thus, the idea of substituting the variable value in the limit is not perfect (it is not a uniform approach), since it is not applicable to all cases (it should be atleast applicable to all non-indeterminate form containing limits, since there are other laws for solving limits with an indeterminate form and combining the two we can create an overall uniform approach). In attempt to create (More precisely "find" since such an approach probably already exists) a uniform approach to solving limits without an indeterminate form (to complete the uniform approach to solving limits), I started with testing whether using the limit laws suffices. I found that it failed when solving limits of the above type. So in a broader sense, my question is: "Is there a uniform approach to solving limits which do not involve an indeterminate form.

My Approach

My first approach was to apply the limit law of quotients : $$\lim_{x\to c}\frac{f(x)}{h(x)}=\frac{\lim_{x\to c}f(x)}{\lim_{x\to c}h(x)}$$ to the limit, taking f(x)=g(x)-g(0) and h(x)=x but the lower limit is 0 so we cannot do so. Then I thought of using the limit law of products : $$\lim_{x\to c}[f(x)h(x)]=\lim_{x\to c}f(x)\lim_{x\to c}h(x)$$ taking f(x)=g(x)-g(0) and h(x)=1/x. Using this I couldn't conclude anything because the limit of h(x) doesn't exist which means that the equality does not hold since the limit laws do not hold.

Any help would be greatly appreciated !

Mayo
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    This reads as if it's a problem that comes from some other problem. Please [edit] the question to tell us that problem (if my guess is right). In any case provide an example of a function $g$. – Ethan Bolker May 13 '25 at 15:29
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    If $g$ is not continuous at $0$ then $\lim_{x \to 0}\frac{g(x)-g(0)}{x}$ can not exist. – Martin R May 13 '25 at 15:49
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    "$\lim_{x \to 0}\frac{g(x)-g(0)}{x}$ where g(x) is discontinuous at 0 but the limit exists" This isn't possible, as @MartinR says. Can you give an example of a function $g$ like this? – Toph May 14 '25 at 06:51
  • @EthanBolker I edited the question, is it fine now? – Mayo May 14 '25 at 08:25
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    I still do not understand the question. The existence of $\lim_{x \to 0}\frac{g(x)-g(0)}{x}$ implies that $g$ is continuous at $0$. Which means that conversely, if $g$ is not continuous at $0$ then the limit does not exist. So what are you trying to compute? – Martin R May 14 '25 at 08:35
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    “Is there a uniform approach to solving every kind of limit” – I do not think so. – Martin R May 14 '25 at 08:36
  • @MartinR The limit doesn't exist but the left hand and right hand limits may... – Mayo May 14 '25 at 10:27
  • @MartinR A uniform approach to solve limits as in something like a flowchart Ex: ...if it is indeterminate form then use lhopital, if lhopital doesn't work use squeeze theorem or expansions ... I specifically want to know a uniform approach for solving limits which do not involve an indeterminate form (as above). – Mayo May 14 '25 at 10:31
  • Can anybody suggest how I Improve the question @MartinR – Mayo May 14 '25 at 12:33
  • @EthanBolker any suggestions? – Mayo May 14 '25 at 12:34
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    In the example in your edit you ask for the limit at $0$ of $(x-2)/x$. That limit does not exist, so it makes no sense to ask for a "rigorous" way to evaluate it. It's not even an indeterminate form. – Ethan Bolker May 14 '25 at 13:29
  • How about the left and right hand limits – Mayo May 14 '25 at 14:17
  • My actual was intention was to be able to evaluate the limit without substituting the function itself ,I.e., to evaluate the limit in general – Mayo May 14 '25 at 15:07
  • @MartinR you can compute this limit in general, but not when g(0) is discontinuous – Conrad12345 May 19 '25 at 20:43

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Your expression is equivalent to the derivative of g(x), at $x=0$. Here's how:

Via definiton, $$g'(a)=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$$

Letting $a=0$,

$$g'(0)=\lim_{x\to 0}\frac{g(x)-g(0)}{x}$$

As you listed above.

Thus, in general, computing $\lim_{x\to 0}\frac{g(x)-g(0)}{x}$ is the same as computing $g'(0)$.

However, if $g(x)$ is discontinuous at $x=0$ (as in your example), then $g(x)$ is not differentiable at $x=0$, and you can't compute g'(0). Alternatively, phrased with the work above -- without continuity, you cannot compute $\lim_{x\to 0}\frac{g(x)-g(0)}{x}$.

Conrad12345
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  • How about LHL and RHL? – Mayo May 18 '25 at 09:59
  • What the values of LHL and the RHL at g(0) is irrelevant -- the above working shows that to find your limit, we need to compute g'(0). To do so, we need the ACTUAL value of g(0), as in the definition of the derivative. This definition makes sense: the derivative tells you the slope of the line at a given point. If the value changes instantly (i.e, a jump discontunity), the slope, rise/run, would give run=0 and be thus be undefined. This is regardless of what g(0) is approached in the LHL or the RHL. – Conrad12345 May 19 '25 at 20:41
  • Alternatively, differentiablity at x=0 means that when you zoom in enough at x=0 the graph starts to look like a straight line. That can't happen if there's a discontinuity, even if the LHL = RHL – Conrad12345 May 19 '25 at 20:50
  • My question clearly asks "how do you calculate the rhl and lhl if limit is not defined" – Mayo May 20 '25 at 01:32
  • I see. To compute g’(0) from the left or right, you still need g(0) to be defined. See https://math.stackexchange.com/questions/1158510/left-hand-derivative-definition – Conrad12345 May 20 '25 at 04:59
  • I know that the LHL and RHL are both infinity (but of different sign) but how do we conclude this rigorously? – Mayo May 21 '25 at 23:26
  • The LHL and the RHL is not infinity — it doesn’t exist. Perhaps I’m misunderstanding your question. How did you arrive that the LHL and the RHL is infinity? The working above is rigorous — take a closer look – Conrad12345 May 21 '25 at 23:35
  • If you work on the extended real number line they do exist and are infinity... – Mayo May 22 '25 at 03:00
  • Do you have an example that you can give? I’m not sure where the confusion is. In the class of limits you gave above, you need continuity, whether you approach from the left or the right. – Conrad12345 May 22 '25 at 06:11
  • No you don't, if you work on the extended real numbers – Mayo May 22 '25 at 06:21