Question

Prove that: lim_x → ∞ [x + (1/x)]^x = e​

Answer 1

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Explanation:

Answer 2

Explanation:

Given Question

Prove that,

`\displaystyle\lim_(x \to \infty )\rm \bigg[1 + {(1)/(x) }\bigg]^(x) = e`

`\red{\large\underline{\sf{Solution-}}}`

Consider,

`\rm :\longmapsto\:\displaystyle\lim_(x \to \infty )\rm \bigg[1 + {(1)/(x) }\bigg]^(x)`

Let assume that

`\rm :\longmapsto\:y = \displaystyle\lim_(x \to \infty )\rm \bigg[1 + {(1)/(x) }\bigg]^(x)`

Taking log on both sides, we get

`\rm :\longmapsto\: log_(e)(y) = \displaystyle\lim_(x \to \infty )\rm log_(e) \bigg[1 + {(1)/(x) }\bigg]^(x)`

We know,

`\red{\rm :\longmapsto\:\boxed{\tt{ log {x}^(y) = y \: logx}}}`

So, using this, we get

`\rm :\longmapsto\: log_(e)(y) = \displaystyle\lim_(x \to \infty )\rm xlog_(e) \bigg[1 + {(1)/(x) }\bigg]`

We know,

`\rm :\longmapsto\:\boxed{\tt{ log_(e)(1 + x) = x - \frac{ {x}^(2) }{2} + \frac{ {x}^(3) }{3} + - - - - }}`

So, using this, we get

`\rm :\longmapsto\: log_(e)(y) = \displaystyle\lim_(x \to \infty )\rm x \bigg[(1)/(x) - {\frac{1}{ {2x}^(2) } + \frac{1}{ {3x}^(3)} - \frac{1}{ {4x}^(4) } + - - - }\bigg]`

So, using this ,we get

`\rm :\longmapsto\: log_(e)(y) = \displaystyle\lim_(x \to \infty )\rm \bigg[1 - {(1)/( 2x ) + \frac{1}{ {3x}^(2)} - \frac{1}{ {4x}^(3) } + - - - }\bigg]`

`\rm :\longmapsto\: log_(e)(y) = 1`

`\bf\implies \:y = e`

`\bf\implies \:\:\displaystyle\lim_(x \to \infty )\rm \bigg[1 + {(1)/(x) }\bigg]^(x) = e`

Hence, Proved

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Additional Information

`\boxed{\tt{ \displaystyle\lim_(x \to 0)\rm (sinx)/(x) = 1}}`

`\boxed{\tt{ \displaystyle\lim_(x \to 0)\rm (tanx)/(x) = 1}}`

`\boxed{\tt{ \displaystyle\lim_(x \to 0)\rm (log(1 + x))/(x) = 1}}`

`\boxed{\tt{ \displaystyle\lim_(x \to 0)\rm \frac{ {e}^(x) - 1}{x} = 1}}`

`\boxed{\tt{ \displaystyle\lim_(x \to 0)\rm \frac{ {a}^(x) - 1}{x} = loga}}`