What is the Limit of a Complex Logarithmic Function in Calculus Homework?

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SUMMARY

The limit of the complex logarithmic function as x approaches 0 is evaluated as follows: lim_{x \to 0} ln\frac{(\sin(1)(17)}{16}. The polynomials in the numerator and denominator simplify to 17 and 16, respectively, when x is substituted with 0. This problem, initially perceived as complex, is resolved through straightforward polynomial reduction and the evaluation of the sine function at a specific value.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with logarithmic functions
  • Knowledge of polynomial simplification
  • Basic understanding of trigonometric functions, specifically sine
NEXT STEPS
  • Study the properties of logarithmic limits in calculus
  • Learn about polynomial long division and simplification techniques
  • Explore the Taylor series expansion for sine functions
  • Practice evaluating limits involving trigonometric functions
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Students in calculus courses, particularly those struggling with limits and logarithmic functions, as well as educators seeking to clarify complex limit evaluations.

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Homework Statement


I have this problem on my calculus homework:

<br /> \lim_{x \to 0} \ln\frac{(\sin(cos(x))(x^5+5x^4+4x^3+17)} {x^6+7x^5+8x^4+9x^3+16})<br />

Homework Equations



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The Attempt at a Solution


I honestly have no idea how to go about this. We really haven't been shown anything like this in class, and the complexity of the problem is quite intimidating. I would be grateful for any sort of help.
 
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Try evaluating it at x=0 first.
 
I got a number like ln.0854, but that was with a calculator which I'm not allowed to use. Not sure how I would do it otherwise
 
Well, every x becomes 0, so your two polynomials reduce to 17 on the top and 16 on the bottom, respectively. Can you see the rest?
 
I have this.. is that all? Is there a way I can evaluate sin(1) without a calculator or do I leave as is?

<br /> <br /> \lim_{x \to 0} \ln\frac{(\sin(1)(17)}{16})<br /> <br />
 
Yes, you have it in perfectly reduced form.
 
That was a lot easier than it looked. I spent so much time trying to make it more complicated then it actually was. Thanks so much for your help.
 

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