I'm sure this is easy for the most of you, but i'd

Format might be weird, but I'll try my best.

As a rule with limits, you can extract constants. So (x^2)/(-x * ln(3)) has the constant [-ln(3)] in the denominator
Rewrite it as (x^2)/(x) * (-1)/(ln(3))
And extract the constant to the front of the limit, and simplify the limit to x {x =\= 0} and viola.
You are left with the second step

Hope this helped.

>>7807305
Cancel out the x and pull out the constant.

>>7807305
It's just a property of limits [eqn] \lim_{ x \to a } c f(x) = c \lim_{x \to a} f(x) [/eqn] Which if you think about it makes a lot of sense. So in this problem [eqn] \lim_{x \to 0^{+}} - \frac { x^2 } { x \ln (3) } = - \frac { 1 }{ \ln(3) } \lim_{x \to 0^{+}} x [/eqn] Where the last equality is just the application of the above property and cancellation of x.

File: Anti-derivative (1).png (2KB, 106x52px) Image search: [Google]

2KB, 106x52px

>>7807305
>>7807332
Thank you so much! while we are at this mind to help me with this anti-derivative?

>>7807341
(ln(2-3x^2))/-6

>>7807351
this is not how numbers work

>>7807341
just substitute x with sqrt(x) to obtain something that looks like the derivative of arcsin

>/sci/ - Teens fresh out of Calc I do Chad's homework because this is the only validation they will ever feel in their sad lives
Pretty good prep for sucking off business grads in your future STEM-monkey careers I suppose.

>>7807326
This is correct. X is a factor on the top and bottom of the fraction, so it can be pulled out as X/X or 1. Then we can pull out -1/ln3 because it is not reliant on X. If it were -X^2/XlnX, it would simplify to -X/lnX and you could not pull out the -1/lnX term.

I hope that makes some sense.

>>7807341

[math] - \frac{1){6} sin^{-1}(\frac{3x^2}{2}) [/math]

[math] - \frac{1){6} sin^{-1}(\frac{3x^2}{2})[/math]

>>7807341

- \frac{1}{6} sin^{-1}(\frac{3x^2}{2})

>>7807341

[math] - \frac{1}{6} sin^{-1}(\frac{3x^2}{2}) [/math]

I'm retarded