- #1
Dannbr
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[tex]\begin{array}{l}
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\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|
\end{array}[/tex]
[tex]\begin{array}{l}
\\
\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|
\end{array}[/tex]
This is as far as i can get I have an answer key which says it simplifies down to..
[tex] = \left| {\frac{x}{3}} \right|[/tex]
could someone help me out with what I am not seeing
Thanks
\\
\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|
\end{array}[/tex]
[tex]\begin{array}{l}
\\
\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|
\end{array}[/tex]
This is as far as i can get I have an answer key which says it simplifies down to..
[tex] = \left| {\frac{x}{3}} \right|[/tex]
could someone help me out with what I am not seeing
Thanks