L'Hôpital's rule — proof using linear approximation, 0 - 0, ∞ - ∞, other indeterminate forms: Question bank

True or false — justify

Whenever a limit gives

\frac{0}{0}

, L'Hôpital's rule will give the answer.

False — it may fail if

lim f^{'} / g^{'}

doesn't exist or loops forever; then you fall back on algebra, Taylor, or the Limits — Definition & Laws. See the

\frac{x + s i n x}{x}

trap.

\frac{0}{0}

is a number equal to

1

because anything over itself is

1

False —

\frac{0}{0}

is an indeterminate form, a question not a value;

\frac{s i n x}{x} \to 1

but

\frac{s i n x}{x ^{2}} \to \infty

, both look like

\frac{0}{0}

. See Indeterminate Forms.

L'Hôpital differentiates the fraction

\frac{f}{g}

using the quotient rule.

False — it differentiates numerator and denominator independently, giving

\frac{f ^{'}}{g ^{'}}

, not

\frac{f ^{'} g - f g ^{'}}{g ^{2}}

. These are entirely different operations.

lim \frac{f ^{'}}{g ^{'}}

does not exist, then

lim \frac{f}{g}

does not exist either.

False — the rule is one-directional.

\frac{x + s i n x}{x} \to 1

but

\frac{1 + c o s x}{1}

oscillates and has no limit; RHS failing tells you nothing.

You may apply L'Hôpital to

\frac{0}{5}

since the top is

0

False — only

\frac{0}{0}

and

\frac{\infty}{\infty}

qualify.

\frac{0}{5} = 0

directly; forcing the rule here gives a wrong or meaningless answer.

0 \cdot \infty

can be fed straight into L'Hôpital.

False — the rule needs a fraction. Rewrite

f \cdot g = \frac{f}{1/ g}

first to make

\frac{0}{0}

\frac{\infty}{\infty}

1^{\infty}

equals

1

because

1

to any power is

1

False — the base only approaches

1

while the exponent grows;

(1 + \frac{1}{x})^{x} \to e

, not

1

. Take

ln

to resolve it.

Repeating L'Hôpital is always legal after the first application.

False — you must re-check that the new fraction is still

\frac{0}{0}

\frac{\infty}{\infty}

before each reapplication. Applying it to a non-indeterminate form is an error.

The linear-approximation derivation is a fully rigorous proof.

False — it secretly assumes

f^{'}, g^{'}

are continuous at

a

. The airtight proof uses the Cauchy Mean Value Theorem, which needs no such assumption.

L'Hôpital works for one-sided limits and for

x \to \pm \infty

True — the theorem is stated for

a

finite,

a \to \pm \infty

, and one-sided limits, provided the hypotheses hold on the relevant side.

Spot the error

lim_{x \to 0} \frac{s i n x}{x} = lim \frac{c o s x}{1} = 1

, so L'Hôpital proves this limit."

Circular — computing

\frac{d}{d x} sin x = cos x

typically uses

lim \frac{s i n x}{x} = 1

. You can't prove the ingredient with the recipe it built.

lim_{x \to \infty} \frac{e ^{x}}{e ^{2 x} + 1}

: differentiate top and bottom repeatedly until it clears."

It never clears — L'Hôpital cycles forever. Use algebra:

\frac{1}{1 + e ^{- 2 x}} \to 1

lim_{x \to π} \frac{s i n x}{x}

: since

sin π = 0

, apply L'Hôpital to get

\frac{c o s x}{1} = - 1

The form is

\frac{0}{π} = 0

, not

\frac{0}{0}

. The denominator doesn't vanish, so the rule doesn't apply; the true value is

0

lim_{x \to 0} \frac{x ^{2}}{x}

\frac{0}{0}

, apply the rule:

\frac{2 x}{1} = 0

The answer

0

is correct, but the setup is wasteful — cancel

x

first to get

x \to 0

. The trap is habitually reaching for L'Hôpital when algebra is trivial.

"For

lim_{x \to 0^{+}} x ln x

, split as

\frac{x}{1/ l n x}

and differentiate."

Legal but disastrous — that derivative is uglier than the original. Choose the split

\frac{l n x}{1/ x}

so differentiation simplifies; the split you pick matters.

lim_{x \to 0} \frac{x ^{2} s i n ( 1/ x )}{x}

\frac{0}{0}

; L'Hôpital gives

\frac{2 x s i n ( 1/ x ) - c o s ( 1/ x )}{1}

, which oscillates, so the limit doesn't exist."

Wrong conclusion — RHS oscillating proves nothing. Cancel:

x sin (1/ x) \to 0

by squeeze. The limit exists and equals

0

Why questions

Why do the constant terms vanish in the linear-approximation derivation?

Because

f (a) = g (a) = 0

(guaranteed by continuity in the

\frac{0}{0}

case), so the tangent lines pass through height zero and only the slope terms survive. See Linear Approximation & Tangent Lines.

Why does the shared factor

(x - a)

cancel, and what does it represent?

Both functions shrink at the same rate

(x - a)

toward the point; that shared "smallness" cancels, leaving the ratio of slopes

\frac{f ^{'} ( a )}{g ^{'} ( a )}

— the ratio of speeds.

Why must we take

ln

to handle

0^{0}

\infty^{0}

1^{\infty}

A variable exponent is invisible to differentiation until

ln

brings it down as a product:

ln (f^{g}) = g ln f

, converting to

0 \cdot \infty

which we can then make a fraction.

Why does the rule require the right-hand limit to exist?

Because the equality only guarantees "if RHS exists then LHS equals it." A non-existent RHS breaks the bridge — the LHS may still exist by other means. See Exponential & Logarithm Growth Rates for cases that need algebra instead.

Why is the Cauchy MVT stronger than the tangent-line picture?

It produces an exact intermediate point

c

with

\frac{f ( x )}{g ( x )} = \frac{f ^{'} ( c )}{g ^{'} ( c )}

using only differentiability — no assumption that

f^{'}, g^{'}

are continuous. See Mean Value Theorem.

Why does

\frac{l n x}{x} \to 0

tell us something about growth rates?

It shows

ln x

grows slower than any positive power of

x

— logarithm loses every race against polynomials. See Exponential & Logarithm Growth Rates.

Why can Taylor series replace L'Hôpital for many limits?

The limit is decided by the leading nonzero terms of top and bottom; their ratio gives the answer, often faster than repeated differentiation. See Taylor Series.

Edge cases

What happens if you apply L'Hôpital to

\frac{\infty}{- \infty}

Legal — the rule covers any

\frac{\pm \infty}{\pm \infty}

; the sign just carries through to the final answer.

Can L'Hôpital handle

\infty - \infty

directly?

No — first combine over a common denominator or factor to produce

\frac{0}{0}

, then apply the rule.

What if

g^{'} (x) = 0

arbitrarily close to

a

The hypothesis "

g^{'} (x) \neq = 0

near

a

" fails, so the theorem doesn't apply; you can't legitimately form

\frac{f ^{'}}{g ^{'}}

in every neighbourhood.

If both

f

and

g

are polynomials vanishing at

a

, do you need L'Hôpital?

No — factor out

(x - a)

and cancel. L'Hôpital works but factoring is cleaner and reveals the multiplicity directly.

Does the rule work when the limit is

\frac{0}{0}

but the derivatives don't exist at

a

itself?

Yes — differentiability is required near

a

(not necessarily at

a

), matching the Cauchy-MVT proof which only samples an interior point

c \to a

What if applying the rule turns

\frac{0}{0}

into a determinate form like

\frac{3}{0 ^{+}}

Then stop — the new form is no longer indeterminate; read off

+ \infty

. Reapplying would be an error since the form isn't

\frac{0}{0}

\frac{\infty}{\infty}

Connections

Parent: L'Hôpital's Rule
Mean Value Theorem — rigor behind the traps.
Indeterminate Forms — why $\frac{0}{0}$ is a question.
Taylor Series — the alternative when L'Hôpital loops.
Exponential & Logarithm Growth Rates — growth-race intuition.
Limits — Definition & Laws — the ground rules.