"Did I Solve the Limit Value Question on Today's Exam Correctly?

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SUMMARY

The limit value question discussed pertains to the evaluation of the limit as x approaches infinity for the expression \(\lim_{x\rightarrow\infty} \cos(\sqrt{2004 + x}) - \cos(\sqrt{x})\). The correct approach involves using the sum-to-product formula and recognizing that as x becomes very large, the expression simplifies to zero. The key steps include approximating \(\sqrt{a+x}\) and analyzing the bounded nature of the trigonometric functions involved.

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KLscilevothma
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This question appeared in today's exam but I think I did it wrongly. I use the "sum to product formula", but I can't find a limit value and say it doesn't exist. Am I correct?

[tex]\lim_{n\rightarrow\infty} cox \sqrt{2004 + x} - cos \sqrt{x}[/tex]
 
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I take it that you are to find the limit when x->inf, not n->inf.
The limit is zero:
Note that sqrt(a+x)=sqrt(x)*sqrt(1+(a/x)) approx. sqrt(x)+1/2(a/sqrt(x)), when a<<x

Hence, we may write the original cosine as cos(sqrt(x)+e),
where e->0 as x->inf.

Using sum-to-product, we have to evaluate the limit of:
cos(sqrt(x))*(cos(e)-1)-sin(e)*sin(sqrt(x)).

Since cos(sqrt(x)), sin(sqrt(x)) are bounded by 1, we see that the whole expression goes to 0.
 
I see. Thanks
 

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