Limit of (1 - cosh(2x)) / 4x^3 + x^2

[username12345]

1. The problem statement, all variables and given/known data

\mathop {\lim }\limits_{x \to 0 } \frac{1 - cosh(2x)}{{4x^3 + x^2}}


2. Relevant equations

Product, sum, quotient laws


3. The attempt at a solution

\mathop {\lim }\limits_{x \to 0 } \frac{1 - cosh(2x)}{{4x^3 + x^2}} =
\mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim cosh(2x)}{{\lim 4 + \lim x^3 + \lim x^2}}
=
\mathop {\lim }\limits_{x \to 0 } \frac{1 - \lim cosh(2x)}{{4 + 0 + 0}}
=
\mathop {\lim }\limits_{x \to 0 } \frac{1 - cosh(\lim 2x)}{{4}}
=
\frac{1 - cosh(0)}{{4}}
= \frac{1 - 1}{{4}}
= 0


However the answer is supposed to be 2.

I was sure the denominator should be 4, but not sure how to get 8 in the numerator :confused:

[Gib Z]

Okay well, You seem to have changed that 4x^3 to 4 + x^3, which is how you got your zero denominator to be 4.

Firstly, do you know \lim_{x\to 0} \frac{ \sinh x}{x} ? That is a good thing to know and is easy to find.

Also, can you express \cosh 2x in terms of sinh^2 x ? That will help.

[username12345]

Okay well, You seem to have changed that 4x^3 to 4 + x^3, which is how you got your zero denominator to be 4.

Firstly, do you know \lim_{x\to 0} \frac{ \sinh x}{x} ? That is a good thing to know and is easy to find.


sinh(0) = 0 so use LHopitals rule :
\lim_{x\to 0} \frac{ \sinh x}{x} = \lim_{x\to 0} \frac{ \cosh x}{1} = \frac{cosh 0}{1} = \frac{1}{1} = 1


Also, can you express \cosh 2x in terms of sinh^2 x ? That will help.
I don't know what you mean, sorry.

[Gib Z]

Ok well, have you seen the identity

\cosh 2x = 2\sinh^2 x +1
?

[username12345]

Ok well, have you seen the identity

\cosh 2x = 2\sinh^2 x +1
?

I tried subbing x = 0 into the above in the original equation and get:

\frac{1 - 2e^2 + 4 - \frac{2}{e^2}}{16x^3 + 4x^2}

Now should I divide each term by x^3?

Can someone please show me the method? I am getting confused now.

[Gib Z]

Directly substitute that identity into your limit, and apply the other limit you worked out.

[username12345]

You mean into this? \mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim cosh(2x)}{{\lim 4 + \lim x^3 + \lim x^2}}

so, \mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim 2 + \lim sinh^2(x+1)}{{\lim 4 + \lim x^3 + \lim x^2}} = \frac{1 - 2 + \lim sinh^2(x+1)}{4}

I can't get that to equal 2.

[jags]

Hey dude apply l-hospital rule when there is 0/0 form after putting x=0 in the original equation , now differentiate the equation not in u/v form differentiate upper equation and lower equation differently until u got the equation not in the form 0/0 after putting the value

[Bohrok]

First use L'Hopital's rule with the original problem and keep using it until you're done. You probably want to use an identity to make differentiation a little easier after one step.

[username12345]

Best answer I can get is 1/2 or -1/2.

Anyway test on this starts in 45 minutes so just hope this question doesn't come up.