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As a beginner in calculus I have always struggled I the area of limits not when I solve higher order thinking questions but just getting the basic idea and the notion of finding limits for a function. It would be a great relief if someone could help me with this query ?

ram kumar
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    To the extent you're saying "Explain limits to me", this request seems rather too broad for a Q&A site; after all, people write entire books about limits. Please refine your question to provide more focus and context. What have you learned so far? What specific obstacle(s) have you encountered? (Examples would be helpful.) Have you done a site search for related questions; and if so, what do you find unsatisfactory about the explanations you found? This can help answerers avoid wasting time explaining things you already understand and can give a sense of how best to provide assistance. – Blue Nov 20 '21 at 15:15
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    As @Blue said, please make your question specific. Right now it is not clear exactly what you didn't understand about the idea of limits. – Laxmi Narayan Bhandari Nov 20 '21 at 15:17
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    Also, I'm a little confused: In a previous question, you ask for strategies for finding the best substitutions for indefinite integration. Integration techniques typically aren't encountered by an "absolute beginner" in calculus. So, again, context would help here. For instance, if you've been self-studying calculus, learning to do various things (differentiation and integration) somewhat mechanically, but have found that you need to fill some holes in your conceptual understanding of limits, say so. – Blue Nov 20 '21 at 15:24
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    In Spivak's Calculus there is an excellent chapter where the author starts with the inuitive idea of a limit and turns progessively this intuition into a rigorous mathematical definition. - Also, a very helpful visual explanation in Thomas' Calculus. – Vince Vickler Nov 21 '21 at 10:36
  • @Blue My answer seems to contradict your assertion that the question is too broad to allow a helpful answer. – user2661923 Nov 23 '21 at 03:34
  • @LaxmiNarayanBhandari Not uncommon for the OP (i.e. original poster) to be vague about what he does not understand. Despite the vagueness of his question, I was able to provide an answer that will help the OP. – user2661923 Nov 23 '21 at 03:36
  • For what it's worth, mathSE reviewers will often be able to provide better assistance if they engage the OP before requiring that the OP provide more information. Often, there is a legitimate roadblock, since you can't provide an answer if the question is low quality. However, there are two considerations: [1] The present question is an interpretation question, so you can't ask the OP to show work. [2] Even with low quality questions, you can give immediate comment/hints to entice the OP into improving the quality of his question. – user2661923 Nov 23 '21 at 03:42

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I disagree with the comments.

Further, since this is an interpretation question, I feel justified in providing an answer even though the OP (i.e. original poster) has shown no work.

I will explain the notion of limits in the simplified world of single variable functions, where both the domain and range of the function is some subset of the Real Numbers. This should give you a reasonable intuitive grasp of the idea behind the limit.

Then, you will have to broaden your intuition to consider functions that have other domains or other ranges.

The first concept to consider is the notion of a neighborhood. The simplest example is to consider a fixed value $a \in \Bbb{R}$. Then, for a small positive value $\delta$, the neighborhood of radius $\delta$ around the value $a$ is regarded as the set of all $x \in \Bbb{R}$ such that $-\delta < (x-a) < \delta.$

Typically, the shorthand expression for this is $|x-a| < \delta.$ Typically, in the definition of a limit, one is concerned with those values of $x$ that are in the neighborhood of radius $\delta$ around $a$, but where $x \neq a.$

Typically, this is expressed as $0 < |x-a| < \delta.$

Then, you have to understand the idea of (for a specific $\epsilon > 0$) the neighborhood of radius $\epsilon$ around some fixed finite value $L$.

Basically, this neighborhood is expressed as the set of all $y$, such that $|y - L| < \epsilon.$

Now, you are ready for the intuitive definition of a limit.

Suppose that you see the assertion that
$\displaystyle \lim_{x \to a} f(x) = L$.

Assigning the variable $y$ to represent $f(x)$, what this assertion signfies, is that for any $\epsilon > 0$ there exists a $\delta > 0$ such that

If $x$ is in a neighborhood of radius $\delta$ around $a$, and $x \neq a$,

Then $y = f(x)$ is in a neighborhood of radius $\epsilon$ around $L$.

More formally, the assertion is written:

$\displaystyle \lim_{x \to a} f(x) = L$ signifies that

For all $\epsilon > 0$, there exists a $\delta > 0$ (where the choice of $\delta$ often depends on the choice of $\epsilon)$

such that $0 < |x - a| < \delta \implies |f(x) - L| < \epsilon.$

As a very simple concrete example, suppose that $f(x) = 2x$, and you are asked to prove that $\displaystyle \lim_{x\to 2} f(x) = 4.$

It turns out that for this particular problem, you can specify $\displaystyle \delta = \frac{\epsilon}{2}.$

Then, if $\displaystyle 0 < |x - 2| < \delta = \frac{\epsilon}{2}$ then you can conclude that

$|f(x) - 4| = |2x - 4| = 2|x - 2| < 2\delta = \epsilon.$

This constitutes a proof that $\displaystyle \lim_{x \to 2} f(x) = 4 ~: ~f(x) = 2x.$

The foundation of the proof was that you were able to identify a relationship between $\delta$ and $\epsilon ~\left(\text{i.e. that} ~\displaystyle \delta = \frac{\epsilon}{2}\right)$ that allowed the required constraint to be satisfied.

user2661923
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