What Is g(x) in the Limit as x Approaches 3?

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SUMMARY

The limit as x approaches 3 for the expression (x+6) / (x^4 - 4x^3 + x^2 + x + 6) equals -1. To rewrite this limit in the form (x-3)g(x), one must manipulate the expression by combining it with 1, resulting in the equation (x+6 + (x^4 - 4x^3 + x^2 + x + 6)) / (x^4 - 4x^3 + x^2 + x + 6) = 0. The next step involves factoring the numerator to identify g(x). This approach is essential for understanding limits involving polynomial expressions in calculus.

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ianq
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Hi,
I'm taking calculus I in college right now and for some reason we stated with limits...We're giving the following limit (sorry, I don't know how to work the board's code to make it look pretty):


lim (x->3) is (x+6) / (x^4 - 4x^3 + x^2 + x + 6) = -1

The prof suggested we rewrite lim (x->3) is (x+6) / (x^4 - 4x^3 + x^2 + x + 6) + 1 = 0 in the form (x-3)g(x) and find g(x). Any idea what g(x) would be and how to find it?
 
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Did you ever think about just plugging in 3 for x and seeing what happens?
 
Yep, I did. I get -1 = -1. But considering it's a class exercise and the prof wants us to rewrite it as (x-3)g(x) I'm clueless...
 
ianq said:
Hi,
I'm taking calculus I in college right now and for some reason we stated with limits...We're giving the following limit (sorry, I don't know how to work the board's code to make it look pretty):


lim (x->3) is (x+6) / (x^4 - 4x^3 + x^2 + x + 6) = -1

The prof suggested we rewrite lim (x->3) is (x+6) / (x^4 - 4x^3 + x^2 + x + 6) + 1 = 0 in the form (x-3)g(x) and find g(x). Any idea what g(x) would be and how to find it?
so do the algebra! What is
\frac{x+6}{x^4- 4x^3+ x^2+ x+ 6}+ 1?

I assume you know that is the same as
\frac{x+6}{x^4- 4x^3+ x^2+ x+ 6}+ \frac{x^4- 4x^3+ x^2+ x+ 6}{x^4- 4x^3+ x^2+ x+ 6}

Add and try to factor the numerator. If you already know one factor, that should be easy!
 

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