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I came across this limit exercise in a book:
$$\lim_{x \to -5}\frac{x^2-25}{x+5}$$

which is then manipulated to $$\lim_{x \to -5}(x-5)=-10$$

But why is just substituting $x=-5$ and letting everything be $0$ considered an invalid move? The ratio of numerator and denominator should remain the same no matter what algebraic manipulation we perform right (multiplying fraction by $(x+5)$)? So how are we getting a different ratio by removing $(x+5)$ from both numerator and denominator?

José Carlos Santos
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user1078
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    No. $0 / 0$ is undefined. You cannot multiply numerator and denominator by $0$ and claim that the result is equal to your original fraction. – K. Jiang Jun 24 '23 at 05:43
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    The idea is that except at $x=-5$ (because then you are actually dividing by $0$), the function is equal to $x-5$, because you can factor the $x+5$ out. Limits are all about the behaviour of functions/sequences near a point in the domain anyway, not actually at it – H. sapiens rex Jun 24 '23 at 05:56
  • @H.sapiensrex according to rules of algebra i should be able to multiply numerator and denominator by same expression right? so in this case if why is removing (x+5) from the numerator and denominator giving a different results than with it? – user1078 Jun 24 '23 at 06:02
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    Which is the text book where the claim is made that it is invalid ? You should include that Page Snippet too. There are various justifications why we can "cancel" $(x+5)$ in Some Cases , Eg 1 : $x$ tends to $-5$ but it is not EQUAL to $-5$ , Eg 2 : Polynomial Considerations ETC. – Prem Jun 24 '23 at 06:36

2 Answers2

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Obviously, you can not substitute $x=-5$ because you get $\frac00$ which is undefined. What you can do is to substitute $x=-5+\epsilon$. Then you get $$\lim_{x \to -5}\frac{x^2-25}{x+5}=\lim_{\epsilon\to 0}\epsilon-10.$$ Now you can substitute $\epsilon=0$ to find the limit.

Bob Dobbs
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The key point is that, by definition of limit, we are interested in the value of the function in a deleted neighborhood of $x=-5$, that is for $x\neq -5$, then we are allowed to perform this step since $x+5\neq 0$

$$\require{cancel}\lim_{x \to -5}\frac{x^2-25}{x+5}=\lim_{x \to -5}\frac{(x-5)\cancel{(x+5)}}{\cancel{(x+5)}}=\lim_{x \to -5} x-5=-10$$

simply concluding by continuity.

Refer also to

user
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  • This was the Crux of my Earlier Comment [[ https://math.stackexchange.com/questions/4724514/why-is-the-value-of-this-limit-equation-without-manipulation-considered-invalid#comment10011039_4724514 ]] , where I asked OP to show which text Book (& which Page) had the claim about invalidity , because it is actually valid in certain Cases. – Prem Jun 24 '23 at 20:08
  • @Prem I think the OP is claiming that it is invalid in the original expression, not in the latter after cancellation. – user Jun 24 '23 at 20:11
  • What you say may be true , though I am not sure whether OP is making the claim or the text book is making the claim or OP is wrongly reading the text book. Basic text Book would have shown that "it is 0/0 , hence more work is required & ETC" & this Question would be unnecessary. – Prem Jun 24 '23 at 20:19