Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

The value of $\underset{x→∞}{\lim}\left(\frac{a_1^{1/x}+a_2^{1/x}+....+a_n^{1/x}}{n}\right),a_i>0$, i = 1, 2, .....n is

Options:

$a_1+a_2+......+a_n$

$e^{a_1+a_2+......+a_n}$

$\frac{a_1+a_2+......+a_n}{n}$

$a_1\,a_2\,a_3...a_n$

Correct Answer:

$a_1\,a_2\,a_3...a_n$

Explanation:

Putting $x=\frac{1}{y}$, we get

L = limit = $\underset{y→0}{\lim}\left(\frac{a_1^y+a_2^y+....+a_n^y}{n}\right)^{n/y}$

$∴\log_eL=\underset{y→0}{\lim}\frac{n}{y}.\log_e\frac{1}{n}(a_1^y+a_2^y+....+a_n^y)$ $(\frac{0}{0})$

$=n\underset{y→0}{\lim}\frac{\frac{n}{n}\left(\frac{a_1^y\log a_1+a_2^y \log a_2+....+a_n^y\log a_n}{a_1^y+a_2^y+....+a_n^y}\right)}{1}=n.\frac{\log(a_1a_2....a_n)}{n}$

$∴\log_L=\log(a_1.a_2. ....a_n)⇒L=a_1.a_2.a_3.....a_n$