lim-x-0-1-mx-1-nx-mn-x-

Question Number 80455 by jagoll last updated on 03/Feb/20

\lim_{x \to 0} {(\frac{1 + m x}{1 - n x})}^{\frac{m n}{x}}

Commented by mr W last updated on 03/Feb/20

\lim_{x \to 0} {(\frac{1 + m x}{1 - n x})}^{\frac{m n}{x}}

= \lim_{x \to 0} {(\frac{\frac{1}{x} + m}{\frac{1}{x} - n})}^{\frac{m n}{x}}

= \lim_{t \to \infty} {[{(\frac{t + m}{t - n})}^{t}]}^{m n}

= \lim_{t \to \infty} {[{(1 + \frac{m + n}{t - n})}^{t}]}^{m n}

= \lim_{t \to \infty} {[{(1 + \frac{m + n}{t - n})}^{t - n} {(1 + \frac{m + n}{t - n})}^{n}]}^{m n}

= \lim_{t \to \infty} {[{(1 + \frac{1}{\frac{t - n}{m + n}})}^{\frac{t - n}{m + n}} {(1 + \frac{m + n}{t - n})}^{\frac{n}{m + n}}]}^{m n (m + n)}

= {[e {(1 + 0)}^{\frac{n}{m + n}}]}^{m n (m + n)}

= e^{m n (m + n)}

Commented by john santu last updated on 03/Feb/20

\lim_{x \to 0} {(1 + \frac{(n + m) x}{1 - n x})}^{\frac{m n}{x}}

\lim_{x \to 0} {(1 + \frac{1}{(\frac{1 - n x}{(n + m) x})})}^{\frac{m n}{x}}

e^{\lim_{x \to 0} (\frac{m n (m + n) x}{(1 - n x) x})} = e^{m n (m + n)} .

Answered by Joel578 last updated on 03/Feb/20

L = \lim_{x \to 0} {{(\frac{1 + m x}{1 - n x})}^{\frac{m n}{x}}}

\ln L = \lim_{x \to 0} {\frac{m n}{x} . \ln (\frac{1 + m x}{1 - n x})}

= \lim {\frac{\ln (1 + m x) - \ln (1 - n x)}{\frac{x}{m n}}}

Using L^{'} h o p i t a l, we get

\ln L = \lim_{x \to 0} {\frac{\frac{m}{1 + m x} - \frac{(- n)}{1 - n x}}{\frac{1}{m n}}} = m n (m + n)

\Rightarrow L = e^{m n (m + n)}

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