If a, b, c, d are positive,then mathop {lim } | vedclass

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(A) We use the standard limit formula $\mathop {\lim }\limits_{u 	o \infty } (1 + \frac{1}{u})^u = e$.Given the expression $\mathop {\lim }\limits_{x

Vedclass · Accepted Answer

$e^{d/b}$

Vedclass · Answer

(A) We use the standard limit formula $\mathop {\lim }\limits_{u 	o \infty } (1 + \frac{1}{u})^u = e$.Given the expression $\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{1}{{a + bx}}} ight)^{c + dx}}$,we rewrite it as:$\mathop {\lim }\limits_{x 	o \infty } {\left[ {\left( {1 + \frac{1}{{a + bx}}} ight)^{a + bx}} ight]^{\frac{c + dx}{a + bx}}}$As $x 	o \infty$,the term $(1 + \frac{1}{a + bx})^{a + bx} 	o e$.Now,we evaluate the limit of the exponent:$\mathop {\lim }\limits_{x 	o \infty } \frac{c + dx}{a + bx} = \mathop {\lim }\limits_{x 	o \infty } \frac{c/x + d}{a/x + b} = \frac{0 + d}{0 + b} = \frac{d}{b}$.Therefore,the limit is $e^{d/b}$.

If $a, b, c, d$ are positive,then $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{a + bx}}} \right)^{c + dx}} = $

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