A calculus problem by أحمد الحلاق

Calculus Level 4

lim ⁡ x → 1 x + 3 x + 7 3 3 x − 2 4 − 4 x − 1 \large \lim_{x\to 1} \dfrac{\sqrt{x+3} \sqrt[3]{x+7} \sqrt[4]{3x-2} - 4}{x-1}

The limit above has a closed form. Find the value of this closed form.

Give your answer to 3 decimal places.


The answer is 3.667.

أحمد الحلاق by أحمد الحلاق , 28, Palestine

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1 solution

L = lim ⁡ x → 1 x + 3 x + 7 3 3 x − 2 4 − 4 x − 1 This is a 0/0 case, so L’H o ˆ pital’s rule applies. = lim ⁡ x → 1 1 2 ( x + 3 ) − 1 2 ( x + 7 ) 1 3 ( 3 x − 2 ) 1 4 + 1 3 ( x + 3 ) 1 2 ( x + 7 ) − 2 3 ( 3 x − 2 ) 1 4 + 3 4 ( x + 3 ) 1 2 ( x + 7 ) 1 3 ( 3 x − 2 ) − 3 4 1 = 1 2 ⋅ 1 2 ⋅ 2 ⋅ 1 + 1 3 ⋅ 2 ⋅ 1 4 ⋅ 1 + 3 4 ⋅ 2 ⋅ 2 ⋅ 1 = 1 2 + 1 6 + 3 = 11 3 ≈ 3.667 \begin{aligned} L & = \lim_{x \to 1} \frac {\sqrt{x+3}\sqrt[3]{x+7}\sqrt[4]{3x-2}-4}{x-1} \quad \quad \small \color{#3D99F6}{\text{This is a 0/0 case, so L'Hôpital's rule applies.}} \\ & = \lim_{x \to 1} \frac {\frac 12 (x+3)^{-\frac 12} (x+7)^\frac 13 (3x-2)^\frac 14 + \frac 13 (x+3)^\frac 12 (x+7)^{-\frac 23} (3x-2)^\frac 14 + \frac 34 (x+3)^\frac 12 (x+7)^\frac 13 (3x-2)^{-\frac 34}}{1} \\ & = \frac 12 \cdot \frac 12 \cdot 2 \cdot 1 + \frac 13 \cdot 2 \cdot \frac 14 \cdot 1 + \frac 34 \cdot 2 \cdot 2 \cdot 1 \\ & = \frac 12 + \frac 16 + 3 = \frac {11}3 \approx \boxed{3.667} \end{aligned}

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I Almost used the same approach , but i like it this way .... \text{I Almost used the same approach , but i like it this way .... }

Let \small \text{Let } f ( x ) = ( x + 3 ) 1 2 ( x + 7 ) 1 3 ( 3 x − 2 ) 1 4 f(x)=(x+3)^{\frac12}(x+7)^{\frac13}(3x-2)^{\frac14}

L = lim ⁡ x → 1 f ( x ) − 4 x − 1 Apply L’Hopitals Rule L = lim ⁡ x → 1 f ′ ( x ) = f ′ ( 1 ) L = \displaystyle\lim_{x \to 1} \dfrac{f(x)-4}{x-1} \quad \quad \small\color{#3D99F6}{\text{Apply L'Hopitals Rule}} \\ L=\displaystyle\lim_{x \to 1} f^{'}(x) = f^{'}(1) \\

Now, Take Log on both sides of our Original Function . We get \text{Now, Take Log on both sides of our Original Function . We get }

ln ⁡ [ f ( x ) ] = 1 2 ⋅ log ⁡ ( x + 3 ) + 1 3 ⋅ log ⁡ ( x + 7 ) + 1 4 ⋅ log ⁡ ( 3 x − 2 ) \ln{[f(x)]}= \dfrac{1}{2}\cdot\log{(x+3)}+\dfrac{1}{3}\cdot\log{(x+7)}+\dfrac{1}{4}\cdot\log{(3x-2)}

Differentiate Both sides w.r.t x \text{Differentiate Both sides w.r.t } x

f ′ ( x ) = f ( x ) ( 1 2 x + 6 + 1 3 x + 21 + 3 12 x − 8 ) L = f ′ ( 1 ) = f ( 1 ) ( 1 8 + 1 24 + 3 4 ) ⟹ L = 1 2 + 1 6 + 3 = 11 3 f^{'}(x)=f(x)\left( \dfrac{1}{2x+6}+\dfrac{1}{3x+21}+\dfrac{3}{12x-8} \right) \\ L=f^{'}(1)=f(1)\left( \dfrac{1}{8}+\dfrac{1}{24}+\dfrac{3}{4} \right) \implies L=\dfrac{1}{2}+\dfrac{1}{6}+3 = \boxed{\dfrac{11}{3}}

Sabhrant Sachan - 4 years, 9 months ago

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