# How to solve this limit?lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1));x>0

Mar 19, 2017

See below.

#### Explanation:

Considering $x \ne 0$

If $\left\mid x \right\mid > 1$

${\lim}_{n \to \infty} \frac{1 + {x}^{n} \left({x}^{2} + 4\right)}{x \left({x}^{n} + 1\right)} = {\lim}_{n \to \infty} {x}^{n} / {x}^{n} \frac{\left(\frac{1}{x} ^ n + \left({x}^{2} + 4\right)\right)}{x \left(1 + \frac{1}{x} ^ n\right)} = \frac{{x}^{2} + 4}{x}$

If $\left\mid x \right\mid < 1$

${\lim}_{n \to \infty} \frac{1 + {x}^{n} \left({x}^{2} + 4\right)}{x \left({x}^{n} + 1\right)} = \frac{1}{x}$

If $x = 1$

${\lim}_{n \to \infty} \frac{1 + {x}^{n} \left({x}^{2} + 4\right)}{x \left({x}^{n} + 1\right)} = 3$

If $x = - 1$

${\lim}_{n \to \infty} \frac{1 + {x}^{n} \left({x}^{2} + 4\right)}{x \left({x}^{n} + 1\right)}$ does not exists

Mar 27, 2017

"The Reqd. Lim.="1/x, if 0 lt x lt 1;
=3, if x=1; and,
$= x + \frac{4}{x} , \mathmr{if} 1 < x .$

#### Explanation:

$\text{Reqd. Lim.} = {\lim}_{n \to \infty} \frac{1 + {x}^{n} \left({x}^{2} + 4\right)}{x \left({x}^{n} + 1\right)}$

$= {\lim}_{n \to \infty} \frac{1 + {x}^{n + 2} + 4 {x}^{n}}{{x}^{n + 1} + x} \ldots \ldots \left(1\right)$

=lim_(n to oo) {x^n(1/x^n+x^2+4)}/{x^n(x+1/x^(n-1))

$= {\lim}_{n \to \infty} \frac{\frac{1}{x} ^ n + {x}^{2} + 4}{x + \frac{1}{x} ^ \left(n - 1\right)} \ldots \ldots \left(2\right) .$

Now, we have to consider 3 Cases :-

Case 1 : $0 < x < 1.$

In this Case , we know that, as $n \to \infty , {x}^{n} \to 0.$

And, $\text{by (1), :., Reqd. Lim.=} \frac{1 + {x}^{2} \left(0\right) + 4 \left(o\right)}{x \left(0\right) + x} = \frac{1}{x} .$

Case 2 : $x = 1.$

$\text{The Reqd. Lim.=} \frac{1 + 1 + 4}{1 + 1} = \frac{6}{2} = 3.$

Case 3 : $x > 1.$

In this Case, $\because 0 < \frac{1}{x} < 1 , \therefore \text{ as } n \to \infty , \frac{1}{x} ^ n \to 0.$

Hence, $\text{by (2), the Reqd. Lim.} = \frac{0 + {x}^{2} + 4}{x + \frac{1}{x} \left(0\right)} = \frac{{x}^{2} + 4}{x} , \mathmr{and} , x + \frac{4}{x} .$

Enjoy Maths.!