Question

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?

Answer 1

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.

Complete step-by-step answer:
To find if a function is convergent at infinity we find the value of the given function at infinity and if it is zero (or a constant)it is a convergent function.
We now find the limit of the function at infinity,
 $ \mathop {\lim }\limits_{n \to \infty } \dfrac{{n + 2}}{{{n^2} - 3}} $
We will now divide and multiply by $ {n^2} $
\[\mathop {\lim }\limits_{n \to \infty } \dfrac{{n + 2}}{{{n^2} - 3}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2}\dfrac{{n + 2}}{{{n^2}}}}}{{{n^2}\dfrac{{{n^2} - 3}}{{{n^2}}}}}\]
Upon solving this we get,
\[\mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2}\dfrac{{n + 2}}{{{n^2}}}}}{{{n^2}\dfrac{{{n^2} - 3}}{{{n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{1}{n} + \dfrac{2}{{{n^2}}}}}{{1 - \dfrac{3}{{{n^2}}}}}\]
Now in the numerator the evaluation of the limit will give the a value near a constant,
\[\dfrac{{\mathop {\lim }\limits_{n \to \infty } \dfrac{2}{{{n^2}}} + 1\dfrac{1}{{\mathop {\lim }\limits_{n \to \infty } n}}}}{{\mathop {\lim }\limits_{n \to \infty } \left( {1 - \dfrac{3}{{{n^2}}}} \right)}} = \dfrac{{\mathop {\lim }\limits_{n \to \infty } \dfrac{2}{{{n^2}}} + \left( 0 \right)}}{{\mathop {\lim }\limits_{n \to \infty } \left( {1 - \dfrac{3}{{{n^2}}}} \right)}}\]
The term \[\mathop {\lim }\limits_{n \to \infty } \dfrac{2}{{{n^2}}}\]
Will also yield zero, therefore the given limit will become zero at the infinity as the numerator is zero. The given function will thus be convergent and hence proved that the function is convergent.

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.