Question

lim x-0 (1+2x)^{(x+3)/x}


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Answer 1

$lim (x \rightarrow0) \: \: {(1 + 2x)}^{ \frac{(x + 3)}{x} }$

$lim (x \rightarrow0) \: \: {(1 + 2x)}^{ (\frac{x }{x} + \frac{3}{x} ) }$

$lim (x \rightarrow0) \: \: {(1 + 2x)}^{ (1+ \frac{3}{x} ) }$

$\tt it \: is \: in \: the \: form :$

$\bf \blue { \boxed{{a}^{b + c} = {a}^{b} + {a}^{c} }}$

$lim (x \rightarrow0) \: \: {(1 + 2x)}^{ 1 } + {(1 + 2x)}^{ \frac{3}{x} }$

$lim (x \rightarrow0) \: \: {(1 + 2x)} + {(1 + 2x)}^{ \frac{3}{x} } ....(1)$

$\tt now \: consider :$

${(1 + 2x)}^{ \frac{3}{x} }$

$lim (x \rightarrow0) \: \: {(1 + 2x)}^{ \frac{3}{x} }$

$\tt applying \: \: \: \purple{ log() } :$

$lim (x \rightarrow0) \: \: log {(1 + 2x)}^{ \frac{3}{x} }$

$lim (x \rightarrow0) \: \: \frac{3}{x} log (1 + 2x)$

$lim (x \rightarrow0) \: \: \frac {3 \: log (1 + 2x)}{x}$

$\tt now \: \purple{difrrentiating} \: both \: numerator\\ \tt and \: denominator:$

$lim (x \rightarrow0) \: \: \frac { \frac{d}{dx} 3 \: log (1 + 2x)}{ \frac{d}{dx} x}$

$\tt we \: know \: that :$

$\blue{ \boxed{\frac{d}{dx} log(a) = \frac{1}{a} }}$

$lim (x \rightarrow0) \: \: \frac { \frac{1}{1 + 2x}}{ 1}$

$lim (x \rightarrow0) \: \: \frac{1}{1 + 2x}$

$lim (x \rightarrow0) \: \: \frac{1}{1 + 2(0)}$

$\therefore lim (x \rightarrow0) \: {(1 + 2x)}^{ \frac{3}{x} } = 1$

$\tt substituting \: it \: in \: (1)$

$lim (x \rightarrow0) \: \: {(1 + 2x)} + {(1 + 2x)}^{ \frac{3}{x} }$

$lim (x \rightarrow0) \: \: {(1 + 2(0))} +1$

$\orange{ \boxed{\therefore \: lim (x \rightarrow0) \: \: {(1 + 2x)}^{ \frac{(x + 3)}{x} } = 2}}$

HOPE THIS HELPS!!