Question

Use L'Hopitals Rule.
Lim x->0 sinh x-x/x^3

Answer 1

\[\lim_{x\to0}\frac{\sinh x-x}{x^3}=\lim_{x\to0}\frac{\cosh x-1}{3x^2}=\lim_{x\to0}\frac{\sinh x}{6x}=\lim_{x\to0}\frac{\cosh x}{6}=\cdots\]

Answer 2

i got that far and i know the answer is 1/6 so i need to figure out how you get 1 on the numerator?

Answer 3

\[\lim_{x\to0}\frac{\sinh x-x}{x^3}=\lim_{x\to0}\frac{\frac{d}{dx}(\sinh x-x)}{\frac{d}{dx}x^3}=\lim_{x\to0}\frac{\frac{d}{dx}\sinh x-\frac{d}{dx}x}{\frac{d}{dx}x^3}\]
What's the derivative of \(x\) with respect to \(x\)?

Answer 4

Or do you mean how do you get the answer \(\dfrac{1}{6}\) ?

Answer 5

yes i don't understand how to get cosh x to be 1

Answer 6

Well, at this point, you can "directly substitute." You're not actually evaluating \(\cosh x\) because the limit part means we're only *approaching* \(x=0\). Since \(\cosh x\) is continuous for \(x\) near 0, you can just evaluate the function at 0, and \(\cosh 0=1\).

Answer 7

Are you asking why \(\cosh x=0\) or why you are able to apply that reasoning?

Answer 8

i guess how you are able to apply the reasoning to get 1/6?

Answer 9

It's for the same reason that you can say \(\displaystyle\lim_{x\to0}x=0\). \(x\) is continuous near \(x=0\). We never actually attain \(x=0\), but we can still say with certainty that \(f(x)=x\to0\) as \(x\to0\). Does that kinda make sense?

Answer 10

yes thank you

Answer 11

You're welcome