Question

Help with Limits

Answer 1

clear the compound fraction by multiplying top and bottom by \(x+1\) would be a good first algebra step

Answer 2

okay so we have this limit:
\[\lim_{x \rightarrow 0}\frac{ \frac{ 1 }{ x+1 }-1 }{ x }\]
is a indetermination of 0/0 on a polinomic limit, we'll have to factor it out, to get rid of the indetermination, let's begin by solving the sustraction on the top of the fraction.
\[\lim_{x \rightarrow 0}\frac{ \frac{ 1-x+1 }{ x+1 } }{ x }\]
easy step, but we have a very nasty fraction here, wich we can traduce into a fractionary division, I'll do it separately from the limit (very useful techinque for solving the limits without making any mistakes):
\[\frac{ \frac{ 1-x+1 }{ x+1 } }{ x } = \frac{ x }{ x+1 }\div \frac{ x }{ 1 }\]
I'll take advantage of the fact that if I flip one of the fractions I can turn the division sign into a multiplication:
\[(\frac{ x }{ x+1 })(\frac{ 1 }{ x })=\frac{ x }{ x ^{2}+x}\]
Since it's a equality I can replace it into the limit and I won't harm it at all:
\[\lim_{x \rightarrow 0}\frac{ x }{ x ^{2}+x }\]
I still have an indetermination, but I can wipe it out by taking common factor x on the denominator:
\[\lim_{x \rightarrow 0}\frac{ x }{ x(x+1) }=\lim_{x \rightarrow 0}\frac{ 1 }{ x+1 }=\frac{ 1 }{ 1 }=1\]