SME
Calculus I · Problem type
L’Hôpital’s rule on (ekx − 1) ÷ (mx)
Use this for “Find the limit of (e⁴ˣ − 1) ÷ (5x) as x → 0.”
Table
An x close to 0 (x)
(ekx − 1) ÷ (mx) there (r)
This is an example. Type your own values to start a new problem.
lim
x
→
0
e
x
−
1
x
=
L
x
An x close to 0
example
r
(ekx − 1) ÷ (mx) there
calculated
Formulas
L
=
k
m
0.6667
=
2
3
r
=
e
k
×
x
− 1
m
×
x
0.6734
=
e
2 × 0.01
− 1
3 × 0.01
Assumptions
•
L’Hôpital’s rule: if the top f and the bottom g both → 0 as x → a, then lim f ÷ g = lim f′ ÷ g′ when that limit exists.
•
Here f(x) = ekx − 1 and g(x) = mx. At x = 0 the top is e⁰ − 1 = 0 and the bottom is 0, the 0/0 form the rule needs.
•
Chain rule: f′(x) = k·ekx, which is k at x = 0, and g′(x) = m. So the limit is k ÷ m.
•
x is a plain number, and the table checks the answer with x on both sides of 0.
Step-by-step
WE KNOW
k = 2, m = 3, x = 0.01
FIND
the limit (L), (ekx − 1) ÷ (mx) there (r)
Step 1 · Find the limit (L)
Use L = k ÷ m
Top
and
bottom
are
both
0
at
x
=
0,
so
take
the
slope
of
each:
k
·
e
kx
is
k
at
0,
and
the
slope
of
mx
is
m
.
L
=
2
3
f
(
x
)
=
e
2
x
−
1
→
f
′(
x
)
=
2
e
2
x
→
f
′(0)
=
2
×
e
0
=
2
g
(
x
)
=
3
x
→
g
′(0)
=
3
lim
x
→
0
of
e
2
x
− 1
3
x
=
2
3
L
=
0.6667
Step 2 · Find (eᵏˣ − 1) ÷ (mx) there (r)
Use r = (ek × x − 1) ÷ (m × x)
Raise
e
to
the
power
kx,
take
away
1,
then
divide
by
mx.
The
closer
x
is
to
0,
the
closer
r
is
to
k
m
.
r
=
e
2 × 0.01
− 1
3 × 0.01
r
=
0.6734
Check
Put the numbers back in. Both sides should match.
0.6667
=
2
3
✓
0.6734
=
e
2 × 0.01
− 1
3 × 0.01
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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