Question bank — Intuitive concept of a limit — table of values, graphical

True — the limit only inspects $x$ near $a$, never $x=a$ itself, so a hole at $a$ leaves the approached value untouched.

False — knowing the value at the point tells you nothing about the approach; a stray dot can sit at $L$ while the curve heads elsewhere. (Agreement of the two is exactly Continuity at a point.)

False — they must also be equal; existing separately isn't enough, as a jump shows left $=1$, right $=2$, both exist, yet disagree.

False as a proof — a table only builds evidence for a forecast; it can never guarantee the trend, since a function could veer at values smaller than any you tabulated.

False — a limit is the target of an endless squeeze; the outputs may approach $2$ forever without any $x\ne a$ landing on it.

False — $f(a)$ can be perfectly defined at a jump; the limit fails because the two sides disagree, not because the point is missing.

False — the limit ignores the lone point $x=a$ entirely, so editing that one value moves nothing about the approach.

True — the limit reads only the surrounding values, which are identical, so the disagreement at $a$ is invisible to it.

The error is treating $\frac{0}{0}$ as a final answer; it's an indeterminate form signalling simplify first — factor and cancel to get $x+1\to 2$.

They agree only for $x\ne 1$; the original is undefined at $1$ while $x+1$ isn't. The cancellation is legal inside the limit precisely because $x\to 1$ keeps $x\ne 1$.

One side is not enough — the right-hand limit could be anything. You must confirm both sides and that they agree before naming the two-sided limit.

Substitution is only a first guess that happens to work for nice (continuous) functions; at holes, jumps, or $0/0$ traps it gives undefined or the wrong value.

There is no number "almost $2$"; the limit is defined as the single value the outputs get arbitrarily close to, and that value is exactly $2$.

That simplification only holds for $x>0$; for $x<0$ it equals $-1$. Left limit $-1\ne$ right limit $+1$, so the two-sided limit does not exist.

A hole removes only the point $(1,2)$, not the approach toward height $2$ from both sides — so the limit is $2$ despite the hole.

Because the whole purpose is to describe well-behaved approach at broken spots; allowing $x=a$ would re-import the very division-by-zero or undefined value we're trying to step around.

A limit names a single destination; if the two approach paths head to different heights there is no one destination, so calling either the "limit" would be arbitrary. (See One-sided limits and limits at infinity.)

A table samples finitely many $x$-values; the true guarantee of "arbitrarily close" needs the Formal epsilon-delta definition of a limit, which handles every nearness at once, not just the rows you wrote.

The $0/0$ comes from a shared factor vanishing on top and bottom; cancelling it exposes the finite value the ratio was really heading toward once the fake cancellation is removed.

The limit reads the trend of nearby outputs (all near $2$); the lone value at $x=1$ is not "near" any other point, so it never enters the squeeze that defines the limit.

"Undefined" means no value; "indeterminate" means the form alone doesn't decide the answer — different $0/0$ expressions can limit to different numbers, so more work is required.

The slope of a curve is a $\frac{0}{0}$ ratio of a vanishing rise over a vanishing run at a single point; only a limit lets us name the value that ratio approaches. (This is The derivative as a limit of difference quotients.)

No — as $x\to 0$ the input $\frac{1}{x}$ races through all values, so the output oscillates endlessly between $-1$ and $1$ and never settles on one destination.

No — "the limit is $+\infty$" describes the behaviour (unbounded growth) but no finite value is approached, so a finite limit does not exist there.

Not as a two-sided limit — you can only take the one-sided limit from the side where the function lives; the missing side has nothing to approach. (See One-sided limits and limits at infinity.)

Not enough — bounding between two different constants leaves room to wander. You need bounds that both approach the same value, which is the Squeeze (Sandwich) Theorem.

Yes — each side individually homes in on its own height, so both one-sided limits exist; the two-sided limit fails only because those two heights differ.

Yes — every nearby output is $0$, so the approached value is $0$ regardless of what happens at $a$; constant functions approach their constant everywhere.

It's not defined — with no other points to approach through, there is no trend to read; the limit concept simply doesn't apply.