Step by step solution to "Evaluate the following limit using the substitution t=x6..."

naivlingr

Answered question

2021-09-16

Evaluate the following limit using the substitution $t = x^{6}$
$lim_{t \to 1} (\frac{\sqrt[3]{t} - 1}{\sqrt{t} - 1})$

Answer & Explanation

Alannej

Skilled2021-09-17Added 104 answers

Fnid the following limit:
$lim_{x \to 1} (\frac{\sqrt[3]{x^{6}} - 1}{\sqrt{x^{6}} - 1})$
Rationalize the expression
$\frac{\sqrt[3]{x^{6}} - 1}{\sqrt{x^{6}} - 1} = \frac{(\sqrt[3]{x^{6}} - 1) (\sqrt{x^{6}} + 1)}{(\sqrt{x^{6}} - 1) (\sqrt{x^{6}} + 1)} = \frac{(\sqrt[3]{x^{6}} - 1) (\sqrt{x^{6}} + 1)}{x^{6} - 1}$
$lim_{x \to 1} \frac{(\sqrt[3]{x^{6}} - 1) (\sqrt{x^{6}} + 1)}{x^{6} - 1}$
Applying the product rule, factor terms whose limits are neither 0 or infinite out of the product
By the product rule,
$lim_{x \to 1} \frac{(\sqrt[3]{x^{6}} - 1) (\sqrt{x^{6}} + 1)}{x^{6} - 1} = (lim_{x \to 1} \frac{\sqrt[3]{x^{6}} - 1}{x^{6} - 1}) lim_{x \to 1} (\sqrt{x^{6}} + 1)$
The limit of a continuous funtion at a point is just its a value there.
$lim_{x \to 1} (\sqrt{x^{6}} + 1) = \sqrt{1^{6}} + 1 = 2$
$2 lim_{x \to 1} \frac{\sqrt[3]{x^{6}} - 1}{x^{6} - 1}$
Rationalize the expression
$\frac{\sqrt[3]{x^{6}} - 1}{x^{6} - 1} = \frac{(\sqrt[3]{x^{6}} - 1) (1 + \sqrt[3]{x^{6}} + {\sqrt[3]{x^{6}}}^{2})}{(x^{6} - 1) (1 + \sqrt[3]{x^{6}} + {\sqrt[3]{x^{6}}}^{2})} = \frac{1}{1 + \sqrt[3]{x^{6}} + {\sqrt[3]{x^{6}}}^{2}}$
$2 lim_{x \to 1} \frac{1}{1 + \sqrt[3]{x^{6}} + {\sqrt[3]{x^{6}}}^{2}}$
The limit of a continuous function at a points is just its value there
$lim_{x \to 1} \frac{1}{1 + \sqrt[3]{x^{6}} + {\sqrt[3]{x^{6}}}^{2}} = \frac{1}{1 + \sqrt[3]{1^{6}} + {\sqrt[3]{1^{6}}}^{2}} = \frac{1}{3}$
$2 \times \frac{1}{3}$
Evaluate 2/3
$\frac{2}{3} = \frac{2}{3}$
Answer
$\frac{2}{3}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Calculus

Didn't find what you were looking for?