Question

if 0 less than a less than B then limit X tends to infinity e b power n + a power and whole to the power 1 by N is equals to​

Answer 1

SOLUTION

TO DETERMINE

$0 < a < b \: \: then$

$\displaystyle \lim_{n \to \infty } \: { \big( {b}^{n} + {a}^{n} \big)}^{ \frac{1}{n} }$

EVALUATION

Here we have to find

$\displaystyle \lim_{n \to \infty } \: { \big( {b}^{n} + {a}^{n} \big)}^{ \frac{1}{n} }$

Let

$\displaystyle x_{n } = \: { \big( {b}^{n} + {a}^{n} \big)}^{ \frac{1}{n} }$

$\displaystyle \implies x_{n } = \: b { \bigg( 1 + { \bigg( \frac{a}{b} \bigg)}^{n} \bigg)}^{ \frac{1}{n} }$

Since

$0 < a < b$

$\displaystyle \: 0 < \frac{a}{b} < 1$

$\displaystyle \implies b < \: b { \bigg( 1 + { \bigg( \frac{a}{b} \bigg)}^{n} \bigg)}^{ \frac{1}{n} } < b. {2}^{ \frac{1}{n} }$

$\displaystyle \implies b < x_{n } < b. {2}^{ \frac{1}{n} } \: \: ....(1)$

Now

$\displaystyle \lim_{n \to \infty } b = b$

$\displaystyle \lim_{n \to \infty } b. {2}^{ \frac{1}{n} } = b$

So by Sandwich Theorem we get from (1)

$\displaystyle \lim_{n \to \infty } x_n = b$

Hence

$\displaystyle \lim_{n \to \infty } \: { \big( {b}^{n} + {a}^{n} \big)}^{ \frac{1}{n} } = b$

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