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

  • #1

Homework Statement



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


Homework Equations



Product, sum, quotient laws


The Attempt at a Solution



[tex]\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
[/tex]

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:
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
4
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 [tex] \lim_{x\to 0} \frac{ \sinh x}{x} [/tex] ? That is a good thing to know and is easy to find.

Also, can you express [tex]\cosh 2x[/tex] in terms of sinh^2 x ? That will help.
 
  • #3
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 [tex] \lim_{x\to 0} \frac{ \sinh x}{x} [/tex] ? That is a good thing to know and is easy to find.
sinh(0) = 0 so use LHopitals rule :
[tex] \lim_{x\to 0} \frac{ \sinh x}{x} = \lim_{x\to 0} \frac{ \cosh x}{1} = \frac{cosh 0}{1} = \frac{1}{1} = 1 [/tex]

Also, can you express [tex]\cosh 2x[/tex] in terms of sinh^2 x ? That will help.
I don't know what you mean, sorry.
 
  • #4
Gib Z
Homework Helper
3,346
4
Ok well, have you seen the identity

[tex]\cosh 2x = 2\sinh^2 x +1[/tex]
?
 
  • #5
Ok well, have you seen the identity

[tex]\cosh 2x = 2\sinh^2 x +1[/tex]
?
I tried subbing x = 0 into the above in the original equation and get:

[tex]\frac{1 - 2e^2 + 4 - \frac{2}{e^2}}{16x^3 + 4x^2}[/tex]

Now should I divide each term by [tex]x^3[/tex]?

Can someone please show me the method? I am getting confused now.
 
  • #6
Gib Z
Homework Helper
3,346
4
Directly substitute that identity into your limit, and apply the other limit you worked out.
 
  • #7
You mean into this? [tex]\mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim cosh(2x)}{{\lim 4 + \lim x^3 + \lim x^2}}[/tex]

so, [tex]\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} [/tex]

I can't get that to equal 2.
 
  • #8
15
0
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
 
  • #9
867
0
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.
 
  • #10
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.
 

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