Question

The limit of a sequence X = (an), an =
1+2n/3n is​

Answer 1

Answer:

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Answer 2

Concept:

To understand this question we need to understand how limits works.

Limit: is defined as a value that a function approaches the output for the given input values.

$\lim_{n \to \ c} f(x)=L$

It is read as “the limit of f of x, as x approaches c equals L”. The “lim” shows limit, and fact that function f(x) approaches the limit L as x approaches c is described by the right arrow.

Given:

The following limit is given

$\lim_{x \to \ 0} X$

$X= a_{n}$ and $a_{n}$$=\frac{1+2n}{3n}$

To find:

The limit of sequence X

Solution:

Now, if

X→0⇒ $a_{n} \to 0$

Then, $\frac{1+2n}{3n} \to 0$

$n \to -\frac{1}{2}$

$\lim_{n \to \ -\frac{1}{2} } \frac{1+2n}{3n}$

⇒$\lim_{n \to \ -\frac{1}{2} } (\frac{1}{3n} +\frac{2n}{3n})$

⇒$\lim_{n \to \ -\frac{1}{2} } \frac{1}{3n} +\lim_{n \to \ -\frac{1}{2} }\frac{2n}{3n}$

⇒$\lim_{n \to \ -\frac{1}{2} } \frac{1}{3n} +\lim_{n \to \ -\frac{1}{2} }\frac{2}{3}$

⇒$\frac{1}{3(-\frac{1}{2}) } +\frac{2}{3}$

⇒$-\frac{2}{3} +\frac{2}{3}$⇒0

Therefore the limit of the sequence is 0.