In question 52 the answer key says it’s b but I feel like it’s d.why would it be each?

There seems to be a lot of misunderstanding of limits in the answers here.  Yes, this limit does exist.

Informally, in order for a limit of a function f(x) to exist at a point P, it has to have the same limit no matter how you get to P.  But the key here is that you have to get there from within the domain of the function.  For regular functions of one variable, there are at most two relevant directions then: from the left and from the right.  Generally we need both of these limits to exist and agree.  In this case though, x=0 is an endpoint of the function's domain.  So, it's totally irrelevant what's going on for x<0.  That has nothing to do with this function or limit at all.  In such a case, we'd only have to consider the limit from the right.

A bit more formally, for the limit of f(x) not to exist at P there have to be two sequences of points {x_n},{y_n}, *taken from the domain of f*, such that

lim (n->∞) x_n = lim (n->∞) y_n = P

but

lim (n->∞) f(x_n) ≠ lim (n->∞) f(y_n).

All such sequences in this example have lim (n->∞) f(x_n) = 1, so we say indeed that lim (x->0) f(x) = 1.
I would put D as well. For the limit to exist at 0, f must be defined in some open neighborhood of 0. Here, it does not seem as if f(x) is defined for any x<0 so the limit at 0 does not exist.

However, there is a notion of "one-sided limits", where x can tend to 0 "from above" (as is the case here) or "from below". So the one-sided limit on the right will exist and equal 1 from the graph. This is presumably what they mean but strictly it's not correct.
I haven't seen a formal answer yet.

Intuitively, imagine a box with height epsilon and width delta. We are going to confine the limit within the box in a very precise way. If you can draw boxes really really flat, and still confine your function to a box of some cleverly chose width, you have a limit.

Formally, if you can prove that for each epsilon, you can find a delta so that all points f of the interval (x-delta,x+delta) are confined to (eta-epsilon,eta+epsilon), then eta is lim f(x).

We get in trouble if we cannot draw such a box, e.g. for sin(1/x) at x=0. But for limit points at the boundary of an open domain there's no problem.
by
B is correct, because as x goes to zero, f(x) approaches 1. Even more specifically, for any values of x < 1 (under the assunption the empty dots are excluded points), a smaller value of x (difference between x and 0) corresponds to a smaller difference between f(0) and f(x), even for an arbitrarily small value of x (>0). You can find more information on that by searching for "Epsilon Delta Criterium".
Also as others mentioned before, normally you would have to examine both approaches to your limit point, but in this specific case, the domain of the function seems to only for positive values of x, so we  can ignore the negative approach to 0.
Limit is the value we get no matter how we reach a point. We tend to see the limit in the domain of a function, so x = 0 + h is in the domain but 0 - h (h > 0, tending to 0) isn't, so we ignore it.

Let's modify the problem to understand better, suppose in the same question you have the domain of $-2,-1$ included as well where the value is 0. Does the limit exist at 0 ?

>!It doesn't! The path is cut from the left side, if limit was to exist f(0) would be continuous.!<

&#x200B;

One tip: If you can draw a  graph without lifting your pencil (from left to right or the other way) that thing is continuous, if it's continuous limit must exist at that point.
by
It might be that it isn't including the whole graph and you're meant to assume that the ends just continue on, which would allow the limit to exist
For the limit of a function going to a point to exist both the left and right limit must exist. Due to only the right limit (x>0) existing here and the left limit (x<0) not existing here, the limit going to 0 does not exist unless your teacher said that if one of the limits exists the limit exists

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