Question

find the limit of the following, without using l'hopitals

Answer 1

\[\lim_{x \rightarrow 0} \frac{\sqrt[3]{1+2x} -1}{ x }\]

Answer 2

You need to determine the functions value as x approaches 0 from each side. (ie x value for .1,.01,.001,.0001 and -.1,-.01,-.001,-.0001.

Answer 3

oh u mean it can only happen numerically?

Answer 4

\[\lim_{x \rightarrow 0}\frac{ \left( \sqrt[3]{1+2x}-1 \right) }{ x }\]
\[=\lim_{x \rightarrow 0}\frac{ \left( 1+2x \right)^{\frac{ 1 }{3 }}-1 }{x }\]
\[[\left( 1+x \right)^n=1+nx+\frac{ n \left( n-1 \right) }{ 2! }x^2+terms~ containing~higher~powers~of~x, \[\left| x \right|\le 1\]]\]
here n=1/3,\[\left| 2x \right|\le 1,\]
calculate and cancel 1 and then take x common and cancel with x of denomintor.

Answer 5

uhmmmm

Answer 6

\[=\lim_{x \rightarrow 0}\frac{1+\frac{ 1 }{ 3 }(2x)+\frac{ \frac{ 1 }{ 3 }(\frac{ 1 }{3 }-1)(2x)^2 }{ 2! }+terms~containing~x^3~and~higher~powers~of~x-1 }{ x }\]
\[=\lim_{x \rightarrow 0}\frac{\frac{ 1 }{3 }(2x)+\frac{ \frac{ 1 }{3 }(\frac{ 1 }{3 }-1) }{ 2! }(2x)^2+terms~ containing~ x^3~and~higher~powers~of~x }{ x }\]
\[=\lim_{x \rightarrow 0}\frac{x \left[ \frac{ 2 }{3 }+\frac{ \frac{ 1 }{ 3 }*\frac{ -2 }{3 }*4x }{ 2! }+terms~containing ~x^2~and~higher~powers~of~x \right] }{x }\]
\[=\frac{ 2 }{3 }+0+0=\frac{ 2 }{3 }\]