Question

Evaluate \lim_ {x rightarrow \infty } \Left [\frac{x-1} {x-2} \right]^ x.​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to \infty } \: {\bigg[\dfrac{x - 1}{x - 2} \bigg]}^{x}$

can be rewritten as

$\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{x - 1}{x - 2} - 1 \bigg]}^{x}$

$\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{x - 1 - (x - 2)}{x - 2}\bigg]}^{x}$

$\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{x - 1 - x + 2}{x - 2}\bigg]}^{x}$

$\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{x - 2}\bigg]}^{x}$

$\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{x - 2}\bigg]}^{(x - 2) \times \dfrac{1}{x - 2} \times x}$

We know that

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

So, using this we get

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \: \bigg[\dfrac{x}{x - 2} \bigg]}$

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \: \bigg[\dfrac{x}{x(1 - \frac{2}{x} )} \bigg]}$

$\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \: \bigg[\dfrac{1}{(1 - \frac{2}{x} )} \bigg]}$

$\rm \: = \: {e}^{1}$

$\rm \: = \:e$

Hence,

$\rm :\longmapsto\: \boxed{ \bf{ \: \displaystyle\lim_{x \to \infty } \: {\bigg[\dfrac{x - 1}{x - 2} \bigg]}^{x} = e}}$

Additional Information :-

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{tan {}^{ - 1} x}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{ {sin}^{ - 1} x}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{ log(1 + x) }{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{{e}^{x} - 1}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{{a}^{x} - 1}{x} = loga}}$

Answer 2

Answer:

Given expression is

\rm :\longmapsto\:\displaystyle\lim_{x \to \infty } \: {\bigg[\dfrac{x - 1}{x - 2} \bigg]}^{x}:⟼

x→∞

lim

[

x−2

x−1

]

x

can be rewritten as

\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{x - 1}{x - 2} - 1 \bigg]}^{x}=

x→∞

lim

[1+

x−2

x−1

−1]

x

\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{x - 1 - (x - 2)}{x - 2}\bigg]}^{x}=

x→∞

lim

[1+

x−2

x−1−(x−2)

]

x

\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{x - 1 - x + 2}{x - 2}\bigg]}^{x}=

x→∞

lim

[1+

x−2

x−1−x+2

]

x

\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{x - 2}\bigg]}^{x}=

x→∞

lim

[1+

x−2

1

]

x

\rm \: = \:\:\displaystyle\lim_{x \to \infty } \: {\bigg[1 + \dfrac{1}{x - 2}\bigg]}^{(x - 2) \times \dfrac{1}{x - 2} \times x}=

x→∞

lim

[1+

x−2

1

]

(x−2)×

x−2

1

×x

We know that

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

x→∞

lim

[1+

x

1

]

x

=e

So, using this we get

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \: \bigg[\dfrac{x}{x - 2} \bigg]}=e

x→∞

lim

[

x−2

x

]

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \: \bigg[\dfrac{x}{x(1 - \frac{2}{x} )} \bigg]}=e

x→∞

lim

[

x(1−

x

2

)

x

]

\rm \: = \: {e}^{\displaystyle\lim_{x \to \infty } \: \bigg[\dfrac{1}{(1 - \frac{2}{x} )} \bigg]}=e

x→∞

lim

[

(1−

x

2

)

1

]

\rm \: = \: {e}^{1}=e

1

\rm \: = \:e=e

Hence,

\rm :\longmapsto\: \boxed{ \bf{ \: \displaystyle\lim_{x \to \infty } \: {\bigg[\dfrac{x - 1}{x - 2} \bigg]}^{x} = e}}:⟼

x→∞

lim

[

x−2

x−1

]

x

=e

Additional Information :-

\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1}}

x→0

lim

x

sinx

=1

\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1}}

x→0

lim

x

tanx

=1

\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{tan {}^{ - 1} x}{x} = 1}}

x→0

lim

x

tan

−1

x

=1

\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{ {sin}^{ - 1} x}{x} = 1}}

x→0

lim

x

sin

−1

x

=1

\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{ log(1 + x) }{x} = 1}}

x→0

lim

x

log(1+x)

=1

\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{{e}^{x} - 1}{x} = 1}}

x→0

lim

x

e

x

−1

=1

\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{{a}^{x} - 1}{x} = loga}}

x→0

lim

x

a

x

−1

=loga