limit as x approaches 8 of (sqrt(7+\sqrt[3]{x)}-3)/(x-8)

⚠Ⓘ (sqrt(7+\sqrt[3]{x)}-3)/(x-8) is read as (sqrt(7+\sqrt[3]{x})-3)/(x-8): the unmatched final parenthesis closes the outer group; write the closing parenthesis after the brace‖A LaTeX root is read as a fractional power: write (expression)^(1/n), with n=2 for a square root

1) Let's evaluate the limit of

\frac{\sqrt{7 + x^{\frac{1}{3}}} - 3}{x - 8}

when

x \to 8

The limit of

x^{\frac{1}{3}}

8^{\frac{1}{3}}

when

x \to 8

⇥1.1) Let's evaluate:

8^{\frac{1}{3}}

⇥

2

The limit of

7 + x^{\frac{1}{3}}

7 + 2

when

x \to 8

⇥1.2) Let's evaluate:

7 + 2

⇥

9

The limit of

\sqrt{7 + x^{\frac{1}{3}}}

\sqrt{9}

when

x \to 8

⇥1.3) Let's evaluate:

\sqrt{9}

⇥

3

The limit of

\sqrt{7 + x^{\frac{1}{3}}} - 3

3 - 3

when

x \to 8

⇥1.4) Let's evaluate:

3 - 3

⇥

0

The limit of

x - 8

8 - 8

when

x \to 8

⇥1.5) Let's evaluate:

8 - 8

⇥

0

For

\frac{\sqrt{7 + x^{\frac{1}{3}}} - 3}{x - 8}

, the limit takes the indeterminate form 0/0 when

x

→

8

2) Let's apply L'Hôpital's rule (limit of a/b = limit of a'/b') by differentiating numerator and denominator:

Let's perform the differentiation:

\frac{d}{d x} (\sqrt{7 + x^{\frac{1}{3}}} - 3)

\frac{d}{d x} (\sqrt{7 + x^{\frac{1}{3}}}) + 0

Apply the differentiation rule:

\frac{d}{d x} (\sqrt{f}) = \frac{d}{d x} (f) \cdot \frac{1}{2 \sqrt{f}}

with f =

7 + x^{\frac{1}{3}}

\frac{d}{d x} (7 + x^{\frac{1}{3}}) \cdot \frac{1}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

ℹ

\frac{d}{d x} (7 + x^{\frac{1}{3}}) \cdot \frac{1}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

\frac{\frac{d}{d x} (7 + x^{\frac{1}{3}})}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

‖

\frac{d}{d x} (7 + x^{\frac{1}{3}})

0 + \frac{d}{d x} (x^{\frac{1}{3}})

\frac{0 + \frac{d}{d x} (x^{\frac{1}{3}})}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

\frac{\frac{1}{3} \cdot x^{(\frac{1}{3} - 1)}}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

ℹ

\frac{1}{3} \cdot x^{(\frac{1}{3} - 1)}

\frac{x^{(\frac{1}{3} - 1)}}{3}

‖

\frac{1}{3} - 1

\frac{1 - 1 \cdot 3}{3}

\frac{\frac{x^{(\frac{1 - 1 \cdot 3}{3})}}{3}}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

\frac{\frac{x^{(\frac{1 - 3}{3})}}{3}}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

\frac{\frac{x^{- \frac{2}{3}}}{3}}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

\frac{\frac{\frac{1}{x^{\frac{2}{3}}}}{3}}{2 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

\frac{1}{x^{\frac{2}{3}}} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{7 + x^{\frac{1}{3}}}}

\frac{1}{x^{\frac{2}{3}}} \cdot \frac{1}{6} \cdot \frac{1}{\sqrt{7 + x^{\frac{1}{3}}}}

\frac{1}{x^{\frac{2}{3}} \cdot 6 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

3) Let's perform the differentiation:

\frac{d}{d x} (x - 8)

1 + 0

1

With the derivatives of the numerator and the denominator, we obtain a new fraction whose limit can be determined:

The limit of

x^{\frac{2}{3}}

8^{\frac{2}{3}}

when

x \to 8

⇥3.1) Let's evaluate:

8^{\frac{2}{3}}

⇥

2^{2}

⇥

4

The limit of

x^{\frac{2}{3}} \cdot 6 \cdot \sqrt{7 + x^{\frac{1}{3}}}

4 \cdot 6 \cdot 3

when

x \to 8

⇥3.2) Let's evaluate:

4 \cdot 6 \cdot 3

⇥

24 \cdot 3

⇥

72

The limit of

\frac{1}{x^{\frac{2}{3}} \cdot 6 \cdot \sqrt{7 + x^{\frac{1}{3}}}}

\frac{1}{72}

when

x \to 8

So the limit of

\frac{\sqrt{7 + x^{\frac{1}{3}}} - 3}{x - 8}

\frac{1}{72}

when

x \to 8

▶Limit:

\frac{1}{72}

4) The numeric value of:

\frac{1}{72}

equals:

▷

\approx 1.3889 \times 10^{- 2}