Find limit of this function!

Calculus Level 2

lim ⁡ x → 0 72 9 x − 24 3 x − 8 1 x + 9 x + 3 x − 1 x 3 = ? \large \lim _{ x\to 0 } \frac { 729^{ x }-243^{ x }-81^{ x }+9^{ x }+3^{ x }-1 }{ x^{ 3 } } = \ ?

6 ( ln ⁡ 3 ) 3 6(\ln3)^3 4 ( ln ⁡ 3 ) 3 4(\ln3)^3 3 ( ln ⁡ 3 ) 3 3(\ln3)^3 5 ( ln ⁡ 3 ) 3 5(\ln3)^3

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Prasenjeet Symon by Prasenjeet Symon , 23, India

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2 solutions

As an alternative, we could first factor the expression to obtain

lim ⁡ x → 0 ( 3 x − 1 ) ( 3 2 x − 1 ) ( 3 3 x − 1 ) x 3 = \lim_{x \rightarrow 0} \dfrac{(3^{x} - 1)(3^{2x} - 1)(3^{3x} - 1)}{x^{3}} =

lim ⁡ x → 0 ( ( 3 x − 1 x ) ( 3 2 x − 1 x ) ( 3 3 x − 1 x ) ) . \lim_{x \rightarrow 0} \left(\left(\dfrac{3^{x} - 1}{x}\right)\left(\dfrac{3^{2x} - 1}{x}\right)\left(\dfrac{3^{3x} - 1}{x}\right)\right).

Now by the basic definition of the derivative we know that lim ⁡ x → 0 3 a x − 1 x \lim_{x \rightarrow 0} \dfrac{3^{ax} - 1}{x} is just the derivative of the function f ( y ) = 3 a y f(y) = 3^{ay} (with respect to y y ) evaluated at y = 0 , y = 0, which comes out to a ln ⁡ ( 3 ) a\ln(3) for real a > 0. a \gt 0.

and so our original limit becomes

( lim ⁡ x → 0 3 x − 1 x ) ( lim ⁡ x → 0 3 2 x − 1 x ) ( lim ⁡ x → 0 3 3 x − 1 x ) = \left(\lim_{x \rightarrow 0} \dfrac{3^{x} - 1}{x} \right) \left( \lim_{x \rightarrow 0} \dfrac{3^{2x} - 1}{x} \right) \left( \lim_{x \rightarrow 0} \dfrac{3^{3x} - 1}{x} \right) =

ln ⁡ ( 3 ) × 2 ln ⁡ ( 3 ) × 3 ln ⁡ ( 3 ) = 6 ( ln ⁡ ( 3 ) ) 3 . \ln(3) \times 2\ln(3) \times 3\ln(3) = \boxed{6(\ln(3))^{3}}.

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Ikkyu San
May 17, 2015

Use "L'Hopital's Rule"

lim ⁡ x → 0 72 9 x − 24 3 x − 8 1 x + 9 x + 3 x − 1 x 3 \displaystyle\lim _{\color{#D61F06}x\to 0}\dfrac{729^\color{#D61F06}x-243^\color{#D61F06}x-81^\color{#D61F06}x+9^\color{#D61F06}x+3^\color{#D61F06}x-1}{\color{#D61F06}x^3}

= lim ⁡ x → 0 ( 3 x ( 3 x − 1 ) 2 ( 4 ⋅ 3 x + 2 ⋅ 3 3 x + 1 + 7 ⋅ 9 x + 1 ) ln ⁡ ( 3 ) 3 x 2 ) =\displaystyle\lim _{\color{#D61F06}x\to 0}\left(\dfrac{3^\color{#D61F06}x\left(3^\color{#D61F06}x-1\right)^2\left(4\cdot 3^\color{#D61F06}x+2\cdot 3^{3\color{#D61F06}x+1}+7\cdot 9^\color{#D61F06}x+1\right)\ln\left(3\right)}{3\color{#D61F06}x^2}\right)

= lim ⁡ x → 0 ( 3 x − 1 ( 3 x − 1 ) 2 ( 4 ⋅ 3 x + 7 ⋅ 3 2 x + 2 ⋅ 3 3 x + 1 + 1 ) ln ⁡ ( 3 ) x 2 ) =\displaystyle\lim _{\color{#D61F06}x\to 0}\left(\dfrac{3^{\color{#D61F06}x-1}\left(3^\color{#D61F06}x-1\right)^2\left(4\cdot 3^\color{#D61F06}x+7\cdot 3^{2\color{#D61F06}x}+2\cdot 3^{3\color{#D61F06}x+1}+1\right)\ln\left(3\right)}{\color{#D61F06}x^2}\right)

= lim ⁡ x → 0 ( 3 x − 1 ( 4 ⋅ 3 x + 4 ⋅ 3 5 x + 2 − 16 ⋅ 2 7 x − 25 ⋅ 8 1 x + 1 ) ln ⁡ 2 ( 3 ) 2 x ) =\displaystyle\lim _{\color{#D61F06}x\to 0}\left(\dfrac{3^{\color{#D61F06}x-1}\left(4\cdot 3^\color{#D61F06}x+4\cdot 3^{5\color{#D61F06}x+2}-16\cdot 27^\color{#D61F06}x-25\cdot 81^\color{#D61F06}x+1\right)\ln ^2\left(3\right)}{2\color{#D61F06}x}\right)

= lim ⁡ x → 0 ( 3 x − 1 ( 4 ⋅ 3 x − 16 ⋅ 3 3 x − 25 ⋅ 3 4 x + 4 ⋅ 3 5 x + 2 + 1 ) ln ⁡ 2 ( 3 ) 2 x ) =\displaystyle\lim _{\color{#D61F06}x\to 0}\left(\dfrac{3^{\color{#D61F06}x-1}\left(4\cdot 3^\color{#D61F06}x-16\cdot 3^{3\color{#D61F06}x}-25\cdot 3^{4\color{#D61F06}x}+4\cdot 3^{5\color{#D61F06}x+2}+1\right)\ln ^2\left(3\right)}{2\color{#D61F06}x}\right)

= lim ⁡ x → 0 ( 3 x − 1 ( 8 ⋅ 3 x + 8 ⋅ 3 5 x + 3 − 64 ⋅ 2 7 x − 125 ⋅ 8 1 x + 1 ) ln ⁡ 3 ( 3 ) 2 ) =\displaystyle\lim _{\color{#D61F06}x\to 0}\left(\dfrac{3^{\color{#D61F06}x-1}\left(8\cdot 3^\color{#D61F06}x+8\cdot 3^{5\color{#D61F06}x+3}-64\cdot 27^\color{#D61F06}x-125\cdot 81^\color{#D61F06}x+1\right)\ln ^3\left(3\right)}{2}\right)

= lim ⁡ x → 0 ( 3 x − 1 ( 8 ⋅ 3 x − 64 ⋅ 3 3 x − 125 ⋅ 3 4 x + 8 ⋅ 3 5 x + 3 + 1 ) ln ⁡ 3 ( 3 ) 2 ) =\displaystyle\lim _{\color{#D61F06}x\to 0}\left(\dfrac{3^{\color{#D61F06}x-1}\left(8\cdot 3^\color{#D61F06}x-64\cdot 3^{3\color{#D61F06}x}-125\cdot 3^{4\color{#D61F06}x}+8\cdot 3^{5\color{#D61F06}x+3}+1\right)\ln ^3\left(3\right)}{2}\right)

= 3 0 − 1 ( 8 ⋅ 3 0 − 64 ⋅ 3 3 ( 0 ) − 125 ⋅ 3 4 ( 0 ) + 8 ⋅ 3 5 ( 0 ) + 3 + 1 ) ln ⁡ 3 ( 3 ) 2 =\dfrac{3^{\color{#D61F06}0-1}\left(8\cdot 3^\color{#D61F06}0-64\cdot 3^{3(\color{#D61F06}0)}-125\cdot 3^{4(\color{#D61F06}0)}+8\cdot 3^{5(\color{#D61F06}0)+3}+1\right)\ln ^3\left(3\right)}{2}

= 6 ln ⁡ 3 ( 3 ) =\boxed{6\ln^3\left(3\right)}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Though a valid approach, your solution is unnecessarily long. Can you find a much simpler approach to this problem? Hints:

  1. You're right about L'hopital Rule.

  2. d d x ( a x ) = ln ⁡ a ⋅ a x \frac d{dx} ( a^x) =\ln a \cdot a^x for a > 0 a> 0 .

  3. How many times do we need to apply L'H? Notice the denominator.

  4. Express in terms of powers of 3 3 .

Bonus question: If the term x 3 x^3 is replaced by x x , find the easiest way to solve this problem without resorting to L'hopital Rule.

Challenge Master: We need to differentiate 3 times.

So it's 1 6 × ( ( ln ⁡ 729 ) 3 − ( ln ⁡ 243 ) 3 − ( ln ⁡ 81 ) 3 + ( ln ⁡ 9 ) 3 + ( ln ⁡ 3 ) 3 ) \frac 16 \times ( ( \ln 729)^3 - ( \ln 243)^3 -( \ln 81 )^3 + ( \ln 9)^3 + ( \ln 3)^3 ) . Apply the logarithmic rule log ⁡ a b c = c log ⁡ a b \log_a b^c = c \log_a b and simplify, we would get the answer 6 ( ln ⁡ 3 ) 3 6 ( \ln 3 )^3 .

Pi Han Goh - 6 years ago

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can you please show a clearer way to simplify the logs into 6ln(3)^3?

Alex Kasantsidis - 6 years ago

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State 729 , 243 , 81 , 9 729,243,81,9 in terms of powers of 3 3 .

Pi Han Goh - 6 years ago

Sorry to bother you, but can you see anything wrong with my posted solution? I believe it is valid to partition the limit like this once we have shown that each component has a limit. Thanks. :)

Brian Charlesworth - 6 years ago

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Haha don't need to apologize here!

It should be a > 0 a>0 because the function g ( x ) = a x − 1 x g(x) = \frac{a^x - 1}{x} is not define for non-integer negative values of a a . Though you don't need to invoke L'hopital rule in the first place, a simple first principle of derivative should have done the trick, that is considering the fact you have factorized it. And as long as you have partitioned your limit by not introducing multiplication / division of 0, then your limit is alright.

Unless you're taking complex numbers into account, I think your solution works fine. I'm surprised that the numerator could be factorized that way in the first place. And I thought this problem couldn't be solved without L'hopital rule. Kudos! +1

Pi Han Goh - 6 years ago

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@Pi Han Goh – Great! Thanks for checking over my solution. Good point about being able to avoid L'Hopital's rule entirely; I've made the appropriate edits. :)

Brian Charlesworth - 6 years ago

@Pi Han Goh – Your statement "It should be a > 0 a>0 because the function g ( x ) = a x − 1 x g(x) = \frac{a^x - 1}{x} is not define for non-integer negative values of a a ." is not quite right in general, I'd say.

It is indeed defined even for a ≤ 0 a\leq 0 . Hell, it's defined even for non-real complex values of a a . The condition of a > 0 a\gt 0 is appropriate if we're dealing exclusively with reals, i.e., considering g ( x ) g(x) to be a strictly real-valued function.

In general, the condition of a > 0 a\gt 0 is not necessary.

The limit lim ⁡ x → 0 a x − 1 x \lim\limits_{x\to 0}\frac{a^x-1}{x} converges and equals ln ⁡ ( a ) \ln(a) for all a ∈ C a\in\Bbb C . You can confirm this using the Maclaurin series for a x a^x which converges for all a ∈ C a\in\Bbb C .

Prasun Biswas - 5 years, 11 months ago

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@Prasun Biswas – I prefer to it keep it real. I don't like imaginary stuff.

Pi Han Goh - 5 years, 11 months ago

[Response to Challenge Master Note]

For the bonus question, we can use an elementary limit which is lim ⁡ x → 0 a x − 1 x = ln ⁡ ( a ) \lim\limits_{x\to 0}\frac{a^x-1}{x}=\ln(a) .

Call the limit in the bonus question as L \mathcal L . We have,

L = lim ⁡ x → 0 72 9 x − 24 3 x − 8 1 x + 9 x + 3 x − 1 x = lim ⁡ x → 0 ( 72 9 x − 1 ) − ( 24 3 x − 1 ) − ( 8 1 x − 1 ) + ( 9 x − 1 ) + ( 3 x − 1 ) x = lim ⁡ x → 0 72 9 x − 1 x − lim ⁡ x → 0 24 3 x − 1 x − lim ⁡ x → 0 8 1 x − 1 x + lim ⁡ x → 0 9 x − 1 x + lim ⁡ x → 0 3 x − 1 x = ln ⁡ ( 729 ) − ln ⁡ ( 243 ) − ln ⁡ ( 81 ) + ln ⁡ ( 9 ) + ln ⁡ ( 3 ) = ln ⁡ ( 3 6 ) − ln ⁡ ( 3 5 ) − ln ⁡ ( 3 4 ) + ln ⁡ ( 3 2 ) + ln ⁡ ( 3 ) = ( 6 − 5 − 4 + 2 + 1 ) ln ⁡ ( 3 ) = 0 \small\begin{aligned}\mathcal{L}&=\lim_{x\to 0}\frac{729^x-243^x-81^x+9^x+3^x-1}{x}\\&=\lim_{x\to 0}\frac{(729^x-1)-(243^x-1)-(81^x-1)+(9^x-1)+(3^x-1)}{x}\\&=\lim_{x\to 0}\frac{729^x-1}{x}-\lim_{x\to 0}\frac{243^x-1}{x}-\lim_{x\to 0}\frac{81^x-1}{x}+\lim_{x\to 0}\frac{9^x-1}{x}+\lim_{x\to 0}\frac{3^x-1}{x}\\&=\ln(729)-\ln(243)-\ln(81)+\ln(9)+\ln(3)\\&=\ln(3^6)-\ln(3^5)-\ln(3^4)+\ln(3^2)+\ln(3)\\&=(6-5-4+2+1)\ln(3)=\boxed{0}\end{aligned}

The proof for the elementary limit used is easily obtained using the Maclaurin series of a x a^x which converges for all a ∈ C a\in\Bbb C .

Prasun Biswas - 5 years, 11 months ago

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