Which Of The Following Limits Is Equal To

You're staring at a limit problem. Think about it: the expression looks messy — maybe a rational function with a zero in the denominator, or a trig function doing something weird near zero, or an indeterminate form like 0/0 or ∞/∞. The question asks: which of the following limits is equal to [something]?

And you're thinking: I just need to know the answer.*

But here's the thing — the answer matters less than the path. So naturally, because on an exam, in a textbook, or in real modeling work, the form* of the limit tells you which tool to reach for. And most students reach for the wrong one.

What Is a Limit, Really?

Skip the epsilon-delta definition for a moment. Practically, a limit asks: as x gets arbitrarily close to some value (or grows without bound), what does the function do?*

It doesn't ask what the function equals* at that point. So naturally, it asks what it approaches*. Because of that, that distinction — value vs. behavior — is where half the confusion lives.

A limit can exist even if the function isn't defined at the point. It can fail to exist even if the function is defined there. And the limit can be a finite number, infinity, or simply not exist at all.

Why the Form Dictates the Method

You see a limit. Your first move shouldn't be "plug it in." Your first move should be: **what form is this?

Direct Substitution Works (Sometimes)

If plugging in the value gives a real number — not 0/0, not ∞/∞, not 0·∞, not ∞ − ∞, not 1^∞, not 0^0, not ∞^0 — you're done. That's the limit.

lim (x→2) (x² + 3x − 1) = 4 + 6 − 1 = 9

Done. Move on.

But the problems that show up on tests* — the ones that ask "which of the following limits is equal to..." — are never this simple. They're designed to trap you on the indeterminate forms.

The Indeterminate Forms You'll Actually See

Form	What It Means	Common Source
0/0	Both numerator and denominator vanish	Rational functions, difference quotients
∞/∞	Both blow up	Rational functions at infinity, growth comparisons
0·∞	One vanishes, one blows up	x ln x as x→0⁺, x sin(1/x)
∞ − ∞	Two infinities cancel	√(x²+x) − x as x→∞
1^∞	Base → 1, exponent → ∞	(1 + 1/x)^x, compound interest limits
0^0	Base → 0, exponent → 0	x^x as x→0⁺
∞^0	Base → ∞, exponent → 0	x^(1/x) as x→∞

Memorize this table. Not because you'll recite it — because recognizing the form instantly* tells you the playbook.

How to Evaluate Limits: The Playbook

1. Algebraic Manipulation (The 0/0 Workhorse)

Rational function giving 0/0? Factor. Cancel. Try again.

lim (x→3) (x² − 9)/(x − 3)
  = lim (x→3) (x−3)(x+3)/(x−3)
  = lim (x→3) (x+3)
  = 6

Rationalizing numerators/denominators works for roots:

lim (x→0) (√(x+4) − 2)/x
  = lim (x→0) (√(x+4) − 2)/x · (√(x+4) + 2)/(√(x+4) + 2)
  = lim (x→0) (x+4 − 4) / [x(√(x+4) + 2)]
  = lim (x→0) 1/(√(x+4) + 2)
  = 1/4

This is bread-and-butter. Now, if you're slow at factoring or rationalizing, you'll be slow at limits. Period.

2. L'Hôpital's Rule (The Calculus Hammer)

Conditions: 0/0 or ∞/∞ form. Differentiable numerator and denominator near the point. Denominator's derivative not zero near the point.

Then:

lim (x→a) f(x)/g(x) = lim (x→a) f'(x)/g'(x)

Provided the right-hand limit exists.

Classic example:

lim (x→0) sin x / x
  = lim (x→0) cos x / 1
  = 1

But — and this is where people lose points — *L'Hôpital only applies to 0/0 and ∞/∞.Now, ** Not 0·∞. Not ∞ − ∞. Not 1^∞. You have to rewrite those first.

lim (x→0⁺) x ln x        (0·∞ form)
  = lim (x→0⁺) ln x / (1/x)   (∞/∞ form — now L'Hôpital works)
  = lim (x→0⁺) (1/x) / (−1/x²)
  = lim (x→0⁺) −x
  = 0

3. Standard Limits You Should Know Cold

These show up constantly* as building blocks:

Limit	Value
lim (x→0) sin x / x	1
lim (x→0) (1 − cos x) / x	0
lim (x→0) (1 − cos x) / x²	1/2
lim (x→0) (e^x − 1) / x	1
lim (x→0) ln(1+x) / x	1
lim (x→∞) (1 + 1/x)^x	e
lim (x→0) (1 + x)^(1/x)	e

Don't re-derive these every time. Recognize them. Build with them. It's one of those things that adds up.

4. Squeeze Theorem (When Oscillation Happens)

−1 ≤ sin(1/x) ≤ 1
−x² ≤ x² sin(1/x) ≤ x²

As x→0, both bounds go to 0. So the middle goes to 0.

Basically the only reliable tool for limits like x sin(1/x) or x² cos(1/x³) — anything where oscillation meets a vanishing factor.

5. Limits at Infinity: Degree Rules

Rational functions as x→∞:

• Degree numerator < degree denominator → 0
• Degree numerator = degree denominator → ratio of leading coefficients
• Degree numerator > degree

…degree numerator > degree denominator → the limit diverges to ±∞, with the sign determined by the ratio of the leading coefficients. Here's a good example:

[ \lim_{x\to\infty}\frac{3x^{4}-2x^{2}+7}{5x^{3}+x-1} =\lim_{x\to\infty}\frac{3x^{4}}{5x^{3}} =\lim_{x\to\infty}\frac{3}{5}x =+\infty . ]

If the leading coefficient of the denominator is negative, the sign flips accordingly. When the degrees differ by more than one, the growth rate is still governed by the highest‑power term; lower‑order terms become negligible in the limit.

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6. Exponential vs. Polynomial Growth

Exponential functions eventually outpace any polynomial. A useful hierarchy to keep in mind (as (x\to\infty)) is

[ \lim_{x\to\infty}\frac{x^{n}}{e^{ax}}=0\quad\text{for any }n\ge0,\ a>0, ] [ \lim_{x\to\infty}\frac{e^{ax}}{x^{n}}=\infty . ]

Because of this, limits such as

[ \lim_{x\to\infty}\frac{x^{5}+3x^{2}}{e^{0.2x}}=0, \qquad \lim_{x\to\infty}\frac{e^{3x}}{x^{100}+1}=\infty ]

Continue exploring with our guides on ethanol is used in the dna isolation process because and write a linear equation given two points.

follow directly from this rule. The same principle applies in reverse for (x\to-\infty) when the exponential base is between 0 and 1 (e.On top of that, g. , (e^{x}\to0)).

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7. Logarithmic Growth

Logarithms grow slower than any positive power of (x):

[ \lim_{x\to\infty}\frac{\ln x}{x^{p}}=0\quad(p>0),\qquad \lim_{x\to\infty}\frac{x^{p}}{\ln x}=\infty . ]

Thus,

[ \lim_{x\to\infty}\frac{\ln(x^{2}+1)}{x}=0, \qquad \lim_{x\to\infty}\frac{x^{0.1}}{\ln x}= \infty . ]

These facts are handy when confronting indeterminate forms like (\infty/\infty) that involve a logarithm paired with a polynomial or exponential term.

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8. Transforming Other Indeterminate Forms

Recall the table of indeterminate forms from the earlier section. Each can be reduced to a (0/0) or (\infty/\infty) situation by algebraic rewriting:

Form	Typical rewrite
(0\cdot\infty)	Write as (\frac{0}{1/\infty}) or (\frac{\infty}{1/0}) → (0/0) or (\infty/\infty)
(\infty-\infty)	Combine over a common denominator or factor out the dominant term
(1^{\infty})	Use the exponential‑log trick: (f(x)^{g(x)}=e^{g(x)\ln f(x)}) → examine (g(x)\ln f(x)) (often (0\cdot\infty))
(0^{0})	Same exponential‑log trick; study (\ln f(x)\cdot g(x))
(\infty^{0})	Again, (e^{g(x)\ln f(x)}) with (g(x)\to0,\ \ln f(x)\to\infty) → (0\cdot\infty)

Example ( (1^{\infty}) ):

[ \lim_{x\to\infty}\left(1+\frac{2}{x}\right)^{3x} = \exp!\left(\lim_{x\to\infty} 3x\ln!\left(1+\frac{2}{x}\right)\right). ]

Since (\ln(1+u)\sim u) for small (u),

[ 3x\ln!\left(1+\frac{2}{x}\right)\sim 3x\cdot\frac{2}{x}=6, ]

so the limit equals (e^{6}).

Example ( (0\cdot\infty) ):

[ \lim_{x\to0^{+}} x\ln x = \lim_{x\to0^{+}} \frac{\ln x}{1/x} \stackrel{\text{L'H}}{=} \lim_{x\to0^{+}} \frac{1/x}{-1/x^{2}} = \lim_{x\to0^{+}} (-x)=0 . ]

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9. Series (Taylor/Maclaurin) Approximations

When direct algebraic manipulation becomes cumbersome, expanding the numerator and denominator as power series around the point of interest often isolates the leading non‑zero term. Simple, but easy to overlook.

For (x\to0),

[ \sin x = x - \frac{x^{3}}{6}+O

Expanding the elementary functions into their Maclaurin series gives a powerful shortcut for limits that would otherwise demand repeated applications of L’Hôpital’s rule.

For the sine function the series begins

[ \sin x = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \cdots . ]

Keeping only the first non‑zero term is enough when (x) is sufficiently close to zero, because the higher‑order pieces become negligible in comparison. Consequently

[ \lim_{x\to0}\frac{\sin x}{x}= \lim_{x\to0}\frac{x-\frac{x^{3}}{6}+O(x^{5})}{x}=1, ]

and the same reasoning yields

[ \lim_{x\to0}\frac{e^{x}-1}{x}= \lim_{x\to0}\frac{x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+O(x^{4})-1}{x}=1. ]

These elementary examples illustrate how the dominant term of a series determines the value of the limit.

The technique extends to more detailed expressions. Consider the limit

[ \lim_{x\to\infty}\left(1+\frac{2}{x}\right)^{3x}. ]

Writing the base as an exponential eliminates the indeterminate form (1^{\infty}):

[ \left(1+\frac{2}{x}\right)^{3x}= \exp!\bigl(3x\ln(1+2/x)\bigr). ]

For large (x), the logarithm can be replaced by its series expansion (\ln(1+u)=u-\frac{u^{2}}{2}+O(u^{3})) with (u=2/x). Substituting gives

[ 3x\ln!\left(1+\frac{2}{x}\right)=3x\left(\frac{2}{x}-\frac{2^{2}}{2x^{2}}+O!\left(\frac{1}{x^{3}}\right)\right)=6-\frac{6}{x}+O!\left(\frac{1}{x^{2}}\right). ]

Hence the exponent tends to (6), and the original limit equals (e^{6}). The same series‑based reasoning works for limits involving logarithms, such as

[ \lim_{x\to\infty}\frac{\ln(1+x)}{x}= \lim_{x\to\infty}\frac{x-\frac{x^{2}}{2}+O(x^{3})}{x}=1, ]

showing that a logarithm grows proportionally to the variable itself when the argument is linear.

Beyond the immediate neighbourhood of a point, series can also be employed to compare growth rates at infinity. Here's a good example: the expansion of the exponential function,

[ e^{ax}=1+ax+\frac{a^{2}x^{2}}{2!}+\frac{a^{3}x^{3}}{3!}+\cdots, ]

demonstrates that any positive power of (x) is eventually dominated by the exponential term, confirming the earlier statement that (\displaystyle\lim_{x\to\infty}\frac{x^{n}}{e^{ax}}=0). Conversely, the reciprocal series for a decaying exponential,

[ e^{-ax}=1-ax+\frac{a^{2}x^{2}}{2!}-\cdots, ]

illustrates that when the exponent is negative the function approaches zero faster than any reciprocal polynomial.

A further refinement is the use of asymptotic notation. When we write (f(x)\sim g(x)) as (x\to\infty), we mean that (\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)}=1). Still, this shorthand allows us to replace a complicated expression by its leading term without sacrificing rigor. Take this: (\sin x\sim x) as (x\to0) and (e^{ax}\sim \frac{e^{ax}}{ax}) as (x\to\infty) (the latter follows from dividing the series by its first non‑constant term).

Summarising the main take‑aways:

• Polynomials are outpaced by exponentials; logarithms are outpaced by any positive power of the variable.
• Indeterminate forms become tractable after rewriting them as ratios or by applying the exponential‑log transformation.
• Power‑series expansions isolate the dominant behaviour near a point or at infinity, turning many limits into straightforward algebraic exercises.
• Asymptotic equivalence provides a concise language for comparing the relative sizes of functions.

By mastering these tools—growth‑rate comparisons, algebraic manipulation of indeterminate forms, and series‑based approximations—students gain a reliable arsenal for tackling virtually any limit encountered in calculus.