Limit Problem with e

[Jason76]

\(\displaystyle \lim x \rightarrow 0 \dfrac{e^{9x} - 1 - 9x}{x^{2}}\)

\(\displaystyle \lim x \rightarrow 0 \dfrac{e^{9(0)} - 1 - 9(0)}{(0)^{2}}\)

\(\displaystyle \lim x \rightarrow 0 \dfrac{e^{(0)} - 1 - 9(0)}{(0)^{2}} = \dfrac{0}{0}\)

\(\displaystyle \lim x \rightarrow 0 \dfrac{9e^{9x} - 9}{2x}\)

\(\displaystyle \lim x \rightarrow 0 \dfrac{9e^{9(0)} - 9}{2(0)} = \dfrac{0}{0}\)

\(\displaystyle \lim x \rightarrow 0 \dfrac{81e^{9x}}{2}\)

\(\displaystyle \lim x \rightarrow 0 \dfrac{81e^{9(0)}}{2} = \dfrac{81}{2} \)

:?: Next move, not getting right answer on the computer.

 

[Jason76]

Never mind that's the right answer. But did I write out the problem correctly? Right notation etc..?

 

[HallsofIvy]

I would recommend that you say that you are using L'Hopital's rule (twice).

 

[DrPhil]

LaTeX hints

\(\displaystyle \lim x \rightarrow 0 \dfrac{e^{9x} - 1 - 9x}{x^{2}}\)

Make {x \rightarrow 0} a SUBSCRIPT of the \lim operator:

\(\displaystyle \lim_{x \rightarrow 0} \dfrac{e^{9x} - 1 - 9x}{x^{2}}\)

Also, to get it all the way under, use \displaystyle:

\(\displaystyle \displaystyle \lim_{x \rightarrow 0} \dfrac{e^{9x} - 1 - 9x}{x^{2}}\)

If you like, you could also put large parentheses around the expression:

\(\displaystyle \displaystyle \lim_{x \rightarrow 0}\left( \dfrac{e^{9x} - 1 - 9x}{x^{2}}\right)\)