Doubt on simplification of function in limits

Question

From long time one doubt always baffles me. let us take a function f(x) for which the limit is indeterminate

Hence we simplify the function f(x) to g(x) for which the limit exists.

Let us take an example

$\lim\limits_{x \to -\frac{\pi}{4}} \frac{1+\sqrt 2 sin\theta}{cos2\theta}$ $(1)$

Initially if you directly substitute the value it gives $\frac{0}{0}$ which is indeterminate form

Hence after we simplify finally we get $\lim\limits_{x \to -\frac{\pi}{4}} \frac{1}{1-\sqrt 2 sin\theta}$ $(2)$

If we take f($-\frac{\pi}{4}$) we get $\frac{1}{2}$. are these pointing to same function?

If yes, Then how come now the simplified function has a limit?

If no, how can we say that $\lim\limits_{x \to -\frac{\pi}{4}} \frac{1+\sqrt 2 sin\theta}{cos2\theta}$ = $\frac{1}{2}$.

If I understand correctly isnt $(2)$ a simplified version of $(1)$? If so why is this discrepancy in limits

Sorry if this sounds stupid or silly,

Kindly clarify my doubt

Answer 1

I think this comes down to a subtle but important distinction between the limits of functions and the values of functions.

To define a limit of a function $f(\theta)$ as $\theta \to-\frac\pi4$, we care about the values of $f(\theta)$ at points close to $\theta=-\frac\pi4,$ but the one point at which we never$^*$ need to evaluate $f(\theta)$ when taking its limit as $\theta \to-\frac\pi4$ is at the point $\theta =-\frac\pi4$ itself.

In general, for a function that is defined in a neighborhood of $\theta =-\frac\pi4$, the limit as $\theta \to-\frac\pi4$ and the value at $\theta =-\frac\pi4$ can be two completely different things. For example, define a function $h$ as follows: $$ h(\theta) = \begin{cases} 17 & \text{$\theta = \frac\pi4 + n\frac\pi2,$ $n$ an integer,} \\ \frac{1+\sqrt 2 \sin\theta}{\cos(2\theta)} & \text{otherwise}. \end{cases} $$ Then $h\left(-\frac\pi4\right) = 17$ but $\lim_{\theta\to-\pi/4 }h(\theta) = \frac12.$

Of course $h(\theta)$ is not continuous at $\theta =-\frac\pi4,$ because continuity means the limit and the value at the point in question must be the same. But if we take a function like $g(\theta) = \frac1{1-\sqrt 2 \sin\theta},$ which is continuous at $\theta =-\frac\pi4,$ then $$\lim_{\theta\to-\pi/4}h(\theta) = \lim_{\theta\to-\pi/4}g(\theta)$$ because the value at $\theta =-\frac\pi4$ itself is irrelevant to the definition of the limit, and $$\lim_{\theta\to-\pi/4}g(\theta) = g\left(-\tfrac\pi4\right)$$ because $g$ is continuous at $\theta =-\frac\pi4.$

All of this would work just as well if $-\frac\pi4$ were not even in the domain of $h,$ as long as every other point in some neighborhood of $-\frac\pi4$ is in the domain. In particular, it works for the function $\frac{1+\sqrt 2 \sin\theta}{\cos(2\theta)},$ whose domain does not include $\theta = -\frac\pi4.$

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$^*$Technically, perhaps I should have written "almost never": at some time or another I recall seeing an example of a textbook in which the definition of the limit of a function at some point in the function's domain required the value of the function at that point to equal the limit; that is, the function had to be continuous at that point. But this seems to be an unusual approach that restricts the applicability of limits too much, and it certainly is not the way I learned calculus.

Answer 2

The subtle point you are missing is that the simplification is (slightly) changing the domain on which the function can be defined by the term used to described it. We are going to talk about real numbers only here, nothing more complex.

Let's take a simple example first. What is $\lim_{x \to 0}\frac{x^2}x$? The function $f(x)=\frac{x^2}x$ can be defined for all $x\in \Bbb R, x \neq 0$. If you simplify it, you get $g(x) = x$, which is a function than can be defined for all $x \in \Bbb R$. So when you get asked to find $\lim_{x \to 0}f(x)$, you realize that $f(x)=g(x)$ for all $x \neq 0$, so you know that $\lim_{x \to 0}f(x) = \lim_{x \to 0}g(x)$.

Now the good thing about $g(x)$ is that it is a continuous function, so it is easy to calculate that limit: $\lim_{x \to 0}g(x) = g(0)$.

In your original example, the term you want to find the limit ($f(\theta)$) of is not defined at $\theta = -\frac{\pi}4$. You simplify $f(\theta)$ to $g(\theta)$, which is equal to $f$ on all points where $f$ is defined, but can be defined on more points. So again you know that $\lim_{\theta \to -\frac{\pi}4}f(\theta) = \lim_{\theta \to -\frac{\pi}4}g(\theta)$.

Again, $g(\theta)$ is a continuous function where it is defined, so you know that $\lim_{\theta \to -\frac{\pi}4}f(\theta) = \lim_{\theta \to -\frac{\pi}4}g(\theta) = g(-\frac{\pi}4) = \frac12$.