Question

$lim \: x \to \infty \: {(1 + \frac{1}{ {x}^{n} }) }^{x} \: for \: n > 0$
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Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{ {x}^{n} } \bigg]}^{x}$

can be rewritten as

$\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{ {x}^{n} } \bigg]}^{ {x}^{n} \times \dfrac{x}{ {x}^{n} } }$

We know,

$\boxed{ \tt{ \: \displaystyle\lim_{x \to \infty } {\bigg[1 + \dfrac{1}{x} \bigg]}^{x} = e \: }}$

So, using this, we get

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty }x \times \dfrac{1}{ {x}^{n} } }$

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } }$

Thus,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{ {x}^{n} } \bigg]}^{x} = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } } \: }}$

Now, 3 cases arises

When n = 1

So, above can be continued as

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } }$

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{1 - 1} } }$

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{0} } }$

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty }1 }$

$\rm \: = \:e$

Case :- 2

When n > 1

So, above limit can be continued as

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } }$

$\rm \: = \: {\bigg[e\bigg]}^{\dfrac{1}{ \infty } }$

$\rm \: = \: {e}^{0}$

$\rm \: = \:1$

Case:- 3

When 0 < n < 1

So, above limit can be continued as

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } }$

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } {x}^{1 - n} }$

$\rm \: = \: {e}^{ \infty }$

$\rm \: = \: \infty$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \: \displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{ {x}^{n} } \bigg]}^{x} = \begin{cases} &\sf{e \: \: when \: n = 1} \\ &\sf{1 \: \: when \: n > 1}\\ &\sf{ \infty \: \: when \:0 < n < 1 } \end{cases}\end{gathered}\end{gathered}$

Answer 2

Answer:

⟼x→∞lim[1+xn1]x

can be rewritten as

\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{ {x}^{n} } \bigg]}^{ {x}^{n} \times \dfrac{x}{ {x}^{n} } }=x→∞lim[1+xn1]xn×xnx

We know,

\boxed{ \tt{ \: \displaystyle\lim_{x \to \infty } {\bigg[1 + \dfrac{1}{x} \bigg]}^{x} = e \: }}x→∞lim[1+x1]x=e

So, using this, we get

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty }x \times \dfrac{1}{ {x}^{n} } }=ex→∞limx×xn1

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } }=ex→∞limxn−11

Thus,

\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{ {x}^{n} } \bigg]}^{x} = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } } \: }}:⟼x→∞lim[1+xn1]x=ex→∞limxn−11

Now, 3 cases arises

When n = 1

So, above can be continued as

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } }=ex→∞limxn−11

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{1 - 1} } }=ex→∞limx1−11

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{0} } }=ex→∞limx01

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty }1 }=ex→∞lim1

\rm \: = \:e=e

Case :- 2

When n > 1

So, above limit can be continued as

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } }=ex→∞limxn−11

\rm \: = \: {\bigg[e\bigg]}^{\dfrac{1}{ \infty } }=[e]∞1

\rm \: = \: {e}^{0}=e0

\rm \: = \:1=1

Case:- 3

When 0 < n < 1

So, above limit can be continued as

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \dfrac{1}{ {x}^{n - 1} } }=ex→∞limxn−11

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } {x}^{1 - n} }=ex→∞limx1−n

\rm \: = \: {e}^{ \infty }=e∞

\rm \: = \: \infty=∞

Hence,

\begin{gathered}\begin{gathered}\begin{gathered}\bf\: \: \displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{ {x}^{n} } \bigg]}^{x} = \begin{cases} &\sf{e \: \: when \: n = 1} \\ &\sf{1 \: \: when \: n > 1}\\ &\sf{ \infty \: \: when \:0 < n < 1 } \end{cases}\end{gathered}\end{gathered}\end{gathered}x→∞lim[1+xn1]x=⎩⎪⎪⎨⎪⎪⎧ewhenn=11whenn>1∞when0<n<1

Step-by-step explanation:

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