- #1

thenewbosco

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the following limit is to be evaluated as x-->infinity

[tex](\frac{x-1}{x+4})^{3x+1}[/tex]

here is the work I've done

taking the natural log:

[tex]3x+1 ln(\frac{x-1}{x+4})[/tex]

to make it an indeterminant form i write it in the following way:

[tex]\frac{ln\frac{x-1}{x+4}}{\frac{1}{3x+1}}[/tex]

applying l'hopital's rule to this yields:

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

simplifying:

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

now noticing that the degree of the numerator is 3 and the denominator is 1, the limit as x--> infinity is infinity.. exponentiating gives e to the infinity...the answer to this is apparently 0 as i found when i graphed the function...where did i go wrong and how to correct this?

[tex](\frac{x-1}{x+4})^{3x+1}[/tex]

here is the work I've done

taking the natural log:

[tex]3x+1 ln(\frac{x-1}{x+4})[/tex]

to make it an indeterminant form i write it in the following way:

[tex]\frac{ln\frac{x-1}{x+4}}{\frac{1}{3x+1}}[/tex]

applying l'hopital's rule to this yields:

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

simplifying:

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

now noticing that the degree of the numerator is 3 and the denominator is 1, the limit as x--> infinity is infinity.. exponentiating gives e to the infinity...the answer to this is apparently 0 as i found when i graphed the function...where did i go wrong and how to correct this?

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