Find the exact value of the following limit: (e^(6x)-6x-1)/(x^2)

Question

chris215 · Answer

I got 18

solomonzelman · Answer

x --> what ?

chris215 · Answer

as x->0

mathmale · Answer

By (e^(6x)-6x-1)/(x^2)

do you mean the following?$$\frac{ e ^{6x}-6x-1 }{ x^2 }?$$

solomonzelman · Answer

$\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{e^{6x}-6x-1}{x^2} }$

chris215 · Answer

yes

solomonzelman · Answer

use L'Hospital's rule.
(it is 0/0 now)

mathmale · Answer

I agree with Solomon:  Use l'H's Rule.
Differentiate the numerator with respect to x.
Differentiate the denominator with respect to x, separately.
Form a fraction out of your results.
Take the limit as x -> 0 of that fraction.

solomonzelman · Answer

differentiate, and if you have questions, ask.

mathmale · Answer

@chris215 :  Share your thoughts, please, and especially share any work you've been able to do on this problem.  Have you worked with l;Hopital's Rule before?

solomonzelman · Answer

If you used L'Hospital's rule twice, then 18 is right

solomonzelman · Answer

well, 18 is just correct...

chris215 · Answer

I differentiated and got d/dx(e^(6 x)-6 x-1) = 6 (e^(6 x)-1)

chris215 · Answer

and x^2= 2x

solomonzelman · Answer

yes, very good

$\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{e^{6x}-6x-1}{x^2} }$

since it is not 0/0 when x=0 is plugged, we differentiate top and bottom.

$\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{\frac{d}{dx}\left[e^{6x}-6x-1\right] }{\frac{d}{dx}\left[x^2\right]} }$  

$\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{6e^{6x}-6}{2x} }$

$\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{3(e^{6x}-1)}{x} }$

0/0 again, and thus differentiate again...

solomonzelman · Answer

you will use the chain rule for the exponent again, and then evaluate the limit.... and you will get 18 - the correct result you already had.

solomonzelman · Answer

((just asking, when you evaluated the limit first, you used an alternative approach... was it graphing? ))

zarkon · Answer

taylor series is the quickest way

mathmale · Answer

Please give @chris215 an opportunity to respond.

solomonzelman · Answer

Zarkon, honestly, that is the first time I have ever heard this from anyone. To me (and to many others), finding L'H'S twice and differentiating is a matter of 1 minute...

chris215 · Answer

yeah I used graphinng

solomonzelman · Answer

but, whatever you say :)

solomonzelman · Answer

Yeah, graphing is good, but not always available (eg tests)

chris215 · Answer

true but thanks so much! that wa very helpful!

solomonzelman · Answer

Anytime:)

Just remember that if you see that when you plug in the variable into the limit you get 0/0 or ∞/∞ then you can differentiate top and bottom (this is L'Hospital's rule)....