/sci/ - Science & Math » Thread #12196728

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Anonymous Mon 05 Oct 19:20:25 2020 No.12196728 View Reply Original Report

Quoted By: >>12196740 >>12196746 >>12196815 >>12196908 >>12196972 >>12197961

How do you solve this?
How are you supposed to apply the hospital rule on a fraction like this example?

Anonymous

Anonymous Mon 05 Oct 2020 19:22:45 No.12196740 Report

Quoted By:

>>12196728
Yes

Anonymous

Anonymous Mon 05 Oct 2020 19:24:34 No.12196746 Report

Quoted By: >>12196754

>>12196728
have you tried expanding by 2+sqrt(x+6) ?

Anonymous

Anonymous Mon 05 Oct 2020 19:27:03 No.12196754 Report

Quoted By: >>12196926

>>12196746
Why? I understand you're trying to get rid of the fraction but how does that happen when there's (x+2) left in the nominator?

Anonymous

Anonymous Mon 05 Oct 2020 19:28:42 No.12196767 Report

Quoted By:

1) Verify that both numerator and denominator evaluate to 0 at x= -2.
2) Differentiate each of the numerator and denominator with respect to x, and evaluate these derivatives at x= -2.
3) If the derivative of the denominator is nonzero and finite, compute the ratio of the derivatives.
4) Otherwise, ...

Anonymous

Anonymous Mon 05 Oct 2020 19:30:14 No.12196776 Report

Quoted By:

multiply and divide by "2 - square root of x + 6"

Anonymous

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Anonymous Mon 05 Oct 2020 19:35:14 No.12196801 Report

Quoted By: >>12196811

This shit doesn't make any sense to me

Anonymous

Anonymous Mon 05 Oct 2020 19:37:36 No.12196808 Report

Quoted By:

Just multiply both nominator and denominator by (2+sqrt(x+6)), m8. Do it on paper then evaluate it.

Anonymous

Anonymous Mon 05 Oct 2020 19:38:12 No.12196811 Report

Quoted By:

>>12196801
Ok nevermind, i think just need more sleep

Anonymous

Anonymous Mon 05 Oct 2020 19:39:04 No.12196815 Report

Quoted By:

>>12196728
multiply and divide by 2+sqrt(x+6)

Anonymous

Anonymous Mon 05 Oct 2020 19:55:22 No.12196871 Report

Quoted By: >>12196888

Taylor series is your friend.

\sqrt{x+6} = 2+\frac{x+2}{4}+O((x+2)^2)

Immediately gives that the value you are looking for is -4.

Anonymous

Anonymous Mon 05 Oct 2020 19:59:20 No.12196888 Report

Quoted By:

>>12196871
Y-you can't do that

Anonymous

Anonymous Mon 05 Oct 2020 20:03:43 No.12196908 Report

Quoted By: >>12196922

>>12196728
when you see a linear expression on the bottom with a radical, clear it by multiplying by its conjugate over its conjugate. the forms play out nicely

Anonymous

Anonymous Mon 05 Oct 2020 20:06:38 No.12196922 Report

Quoted By:

>>12196908
>its conjugate
Or the product of its conjugates :)

Anonymous

Anonymous Mon 05 Oct 2020 20:07:15 No.12196926 Report

Quoted By:

>>12196754
no, you get a - 2 - x in the denominator, which is exactly -(x + 2). You can then divide that out, and you get that this is exactly

$\lim_{x -\to -2} - \left( 2 + \sqrt{x + 6} \right) = 0$

Anonymous

Anonymous Mon 05 Oct 2020 20:18:48 No.12196972 Report

Quoted By: >>12197441

>>12196728
just set x to -2 lmao, it always works out nicely for simple functions

Anonymous

Anonymous Mon 05 Oct 2020 22:50:58 No.12197441 Report

Quoted By:

>>12196972
Brainlet tier post
You get division by zero here

Anonymous

Anonymous Mon 05 Oct 2020 23:42:13 No.12197619 Report

Quoted By:

$\lim_{x \to -2} \frac{\frac{d}{dx}\left( x+2\right)}{\frac{d}{dx}\left(2-\sqrt{x+6}\right)} = -4$

Anonymous

Anonymous Tue 06 Oct 2020 01:18:55 No.12197961 Report

Quoted By:

>>12196728
>hospital
L'Hôpital

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