Question

Limit

Answer 1

\[\lim_{h \rightarrow 0} \frac{ (3+h)^{-1}-3^{-1} }{ h }\]

Answer 2

\[\lim_{h \rightarrow 0}\frac{ \left( 3+h \right)^{-1}-3^{-1} }{ h }\]
\[=\lim_{h \rightarrow 0}\frac{ 3^{-1}\left( 1+\frac{ h }{ 3 } \right)^{-1}-3^{-1} }{ h }\]
\[=3^{-1}\lim_{h \rightarrow 0}\frac{ \left( 1+\frac{ h }{ 3 } \right)^{-1}-1 }{ h }\]
\[=\frac{ 1 }{ 3 }\lim_{h \rightarrow 0}\frac{ 1+\left( -1 \right)\frac{ h}{ 3 }+terms ~containing~h^2~and~higher~powers~of~h-1 }{ h }\]
\[=\frac{ 1 }{ 3 }\lim_{h \rightarrow 0}\frac{ h \left( \frac{ -1 }{ 3 } +~terms ~containing~h ~and~higher~ powers~of~h\right) }{ h }\]
=?

Answer 3

I thought you'd just simply reciprocate the terms and then multiply it to H

Answer 4

btw thanks :)

Answer 5

yw