Question

How would we solve this:
\[\lim_{x \rightarrow 0} \frac{\sinh x - \sin x}{x^{3}}\]

Without using a graphing tool.

Answer 1

._.

Answer 2

Our limit is giving us the indeterminate form 0/0.
So I guess we could apply L'Hopital's Rule.
Have you learned about this yet?

Answer 3

Nope

I noticed it was giving 0/0 also

but I thought about rationalizing the numerator, I don't think we can do that though

Answer 4

Hmm :\

Answer 5

yeah, which is why I'm confused as well xD

We can break apart the fraction and get

\[\lim_{x \rightarrow 0} \frac{\sinh x - \sin x}{x^{3}}\]
\[\lim_{x \rightarrow 0}( \frac{\sinh x}{x^3} - \frac{\sin x}{x^{3}})\]

and then

\[\lim_{x \rightarrow 0} \frac{\sinh x}{x^3} -\lim_{x \rightarrow 0} \frac{\sin x}{x^{3}}\]
\[\lim_{x \rightarrow 0} \frac{\sinh x}{x^3} -\lim_{x \rightarrow 0} (\frac{\sin x}{x}\cdot \frac{1}{x^2})\]


mhmmmm

Answer 6

Ya I can't seem to figure out the algebraic approach for this one...
I know L'Hopital will work.
You have to apply it three times though.\[\large\rm ~~~\lim_{x\to0}\frac{\sinh x-\sin x}{x^3}\quad=\quad \lim_{x\to0}\frac{\cosh x-\cos x}{3x^2}\]\[\large\rm =\lim_{x\to0}\frac{\sinh x+\sin x}{6x}\quad=\quad \lim_{x\to0}\frac{\cosh x+\cos x}{6}\]
Differentiating the numerator and denominator separately.
Checking to make sure you still have your indeterminate form 0/0 after each application.

Then finally you can plug in x=0 once you've gotten rid of the x in the denominator.
Leading to 2/6 or 1/3.

Answer 7

L'Hopital's rule is that easy? ;o

woah, thanks zeppy o;