can anyone help me solve the lim of lnx/x-1 as x goes to 1
\[\lim_{x \rightarrow 1}\ln x \div x-1\]
\[\lim_{x\rightarrow 1} (\frac{\ln x}{x-1})?\]
coincidentally...if you substitute x = 1 right now..you'll get 0/0 <--you can use l'hospital
pls tell me you know l'hospital....
i hate limits that dont use l'hospital lol
yes I know
and yes that is the right equation
ahh wonderful!
take the derivative of numerator and denominator \[\frac{\frac{1}{x}}{1} \implies \frac 1x\] now sub in \[\lim_{x \rightarrow 1}\frac{1}{1} \implies 1\]
do you get that?
@Zarkon feel free to point out if i made a mistake...limits is not my field of expertise i admit
yes
nice!
trust me I'm just learning this stuff and I only have 3 weeks to get this Calculus class over.
l'hospital is the only topic i remember from limits lol
lol
i suggest you practice your derivatives though...i believe that makes up 3/4 of the calculus I curriculum so if you master derivatives..you're good...and it'll make it easier for you to do l'hospital
ok thanks :)
can someone help me slove this attachment below.
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