- #1

- 10

- 0

Check it out!

Find the limit of x as it approaches a for

y=[[(2*a^3*x-x^4)^.5]-[a^3(a^2x)^.5]]/(a-(ax^3)^.25)

Unfortunately, I can't get rid of the indeterminate form! Ideas/help?

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- Thread starter mathwiz123
- Start date

- #1

- 10

- 0

Check it out!

Find the limit of x as it approaches a for

y=[[(2*a^3*x-x^4)^.5]-[a^3(a^2x)^.5]]/(a-(ax^3)^.25)

Unfortunately, I can't get rid of the indeterminate form! Ideas/help?

- #2

- 261

- 2

[tex]

y = \frac{\left ( \sqrt{2a^3x-x^4}-a^3\sqrt{a^{2x}} \right )}{a - \sqrt{ax^3}}

[/tex]

- #3

- 10

- 0

The diviser should be [a-(ax^3)^.25]

- #4

- 1,444

- 2

[tex]

y = \frac{\left ( \sqrt{2a^3x-x^4}-a^3\sqrt{a^{2x}} \right )}{a - \sqrt{\sqrt{ax^3}}}

[/tex]

y = \frac{\left ( \sqrt{2a^3x-x^4}-a^3\sqrt{a^{2x}} \right )}{a - \sqrt{\sqrt{ax^3}}}

[/tex]

- #5

- 10

- 0

and instead of a^3(a*a*x), it should be a(a*a*x)^(1/3)

- #6

- 1,444

- 2

i wonder if flat out differentiaon would just do it...

maybe you rationalize the numerator?

maybe you rationalize the numerator?

- #7

- 10

- 0

- #8

- 1,444

- 2

https://www.physicsforums.com/misc/howtolatex.pdfmathwiz123 said:

- #9

- 1,235

- 1

[tex]

y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2x}} \right )}{a - \sqrt[4]{ax^3}}

[/tex]

y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2x}} \right )}{a - \sqrt[4]{ax^3}}

[/tex]

- #10

- 10

- 0

[tex]y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2}x} \right )}{a - \sqrt[4]{ax^3}}[\tex]

- #11

- 1,235

- 1

[tex] y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2}x} \right )}{a - \sqrt[4]{ax^3}} [/tex]

- #12

- 10

- 0

Yes! exactly. Thanks for the help.

- #13

- 13,110

- 663

[tex] a^4 -ax^3 [/tex]

which is easy to differentiate when using the 'H^opital rule.

Daniel.

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