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# Using the definition of convergence, how do you prove that the sequence  $\lim \dfrac{{n + 2}}{{{n^2} - 3}} = 0$ converges from $n = 1$ to infinity?

Last updated date: 12th Jul 2024
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Hint: We will first define the convergence of sequences. The convergence of sequence is when a value of a function keeps on decreasing and reaches a constant value or zero at infinity. Say a function gives value of $2$ at $x = 1$ and value of $1$ at $x = 1000$ and then gives value of $0$ at infinity it is said to be convergent .Thus we will first find the limit of the given function at infinity and if it is zero or a constant then this function will be a convergent function. Since we have to prove we already know that it is a convergent function.

$\mathop {\lim }\limits_{n \to \infty } \dfrac{{n + 2}}{{{n^2} - 3}}$
We will now divide and multiply by ${n^2}$
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{n + 2}}{{{n^2} - 3}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2}\dfrac{{n + 2}}{{{n^2}}}}}{{{n^2}\dfrac{{{n^2} - 3}}{{{n^2}}}}}$
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2}\dfrac{{n + 2}}{{{n^2}}}}}{{{n^2}\dfrac{{{n^2} - 3}}{{{n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{1}{n} + \dfrac{2}{{{n^2}}}}}{{1 - \dfrac{3}{{{n^2}}}}}$
$\dfrac{{\mathop {\lim }\limits_{n \to \infty } \dfrac{2}{{{n^2}}} + 1\dfrac{1}{{\mathop {\lim }\limits_{n \to \infty } n}}}}{{\mathop {\lim }\limits_{n \to \infty } \left( {1 - \dfrac{3}{{{n^2}}}} \right)}} = \dfrac{{\mathop {\lim }\limits_{n \to \infty } \dfrac{2}{{{n^2}}} + \left( 0 \right)}}{{\mathop {\lim }\limits_{n \to \infty } \left( {1 - \dfrac{3}{{{n^2}}}} \right)}}$
The term $\mathop {\lim }\limits_{n \to \infty } \dfrac{2}{{{n^2}}}$
Note: The opposite property of a convergent function is the divergent function. In these types of functions instead of decreasing as the $x$ approaches infinity the function gives exceedingly large value. The limit of such a function on infinity will be infinity i.e. they grow without bounds.