can anyone count ts limit for me

[vague lagoon]

pls help

[gilded lion]

Is that supposed to be an x in the numerator of the -2?

Is it 1 -(2/x)?

@vague lagoon

[wicked garnet]

hi, theres a very easy solution for this problem. if you separate the -2/x and lnx terms, you can analyse them individually for x approaches 0. the term lnx approaches -infinity while -2/x diverges to -infinity and +infinity at the same time. since one term diverges, and the whole expression is not an indeterminate form, the limit must diverge.

[pliant hearth]

so is it this? $lim_{x\to0^+} 1 - \frac{2}{x} ln(x)=?$

[shut fractal]

twitch Tamplr

[pliant hearth]

if so then take $v\to0^+$ so that $2/v\to\infty$ and since $ln(v)$ is well known to be $ln(v)=-\infty$ for small v then $lim_{x\to0^+} 1 - \frac{2}{x} ln(x)=1 - \frac{2}{v} ln(v)$ now also take n instead of $\infty$ and find that $1-\frac{2}{v}ln{v}=1-n(-n)=1+n$ and since $n\to\infty$ then we obtain $1+n=\infty$ hence $lim_{x\to0^+} 1 - \frac{2}{x} ln(x)=\infty$

[shut fractal]

twitch Tamplr

[pliant hearth]

make sense?

[vague lagoon]

yall fkd me up

im stupid at math

[vague lagoon]

not really

[vague lagoon]

i dont get it

[vague lagoon]

yes, i guess

[pliant hearth]

Damn😔