Question

Solve this ! Topic: limits​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\lim_{x \to a} \frac{ {a}^{x} - {x}^{a} }{ {x}^{a} - {a}^{a} } = - 1$

$\large\underline{\sf{To\:Find - }}$

The value of a

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

Given that,

$\rm :\longmapsto\:\displaystyle\lim_{x \to a} \frac{ {a}^{x} - {x}^{a} }{ {x}^{a} - {a}^{a} } = - 1$

If we put directly x = a, we get indeterminant form.

So, we use here L - Hospital Rule to evaluate this limit.

$\rm :\longmapsto\:\displaystyle\lim_{x \to a} \frac{\dfrac{d}{dx}( {a}^{x} - {x}^{a} )}{\dfrac{d}{dx}( {x}^{a} - {a}^{a} )} = - 1$

We know,

$\boxed{ \rm{ \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v}}$

So, using this, we get

$\rm :\longmapsto\:\displaystyle\lim_{x \to a} \frac{\dfrac{d}{dx} {a}^{x} - \dfrac{d}{dx}{x}^{a} }{ \dfrac{d}{dx}{x}^{a} - \dfrac{d}{dx}{a}^{a} } = - 1$

We know,

$\boxed{ \rm{ \dfrac{d}{dx} {a}^{x} = {a}^{x}loga}}$

$\boxed{ \rm{ \dfrac{d}{dx}k = 0}}$

$\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}$

So, using these results, we get

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to a} \frac{ {a}^{x} loga - {ax}^{a - 1} }{ {ax}^{a - 1} - 0} = - 1}$

${\rm :\longmapsto\:\displaystyle\lim_{x \to a} \frac{ {a}^{x} loga - {ax}^{a - 1} }{ {ax}^{a - 1}} = - 1}$

${\rm :\longmapsto\: \dfrac{ {a}^{a} loga - {a(a)}^{a - 1} }{ {a(a)}^{a - 1}} = - 1}$

${\rm :\longmapsto\: \dfrac{ {a}^{a} loga - {a}^{a } }{ {a}^{a}} = - 1}$

${\rm :\longmapsto\: \dfrac{ {a}^{a}( loga - 1) }{ {a}^{a}} = - 1}$

$\rm :\longmapsto\:loga - 1 = - 1$

$\rm :\longmapsto\:loga = 0$

$\rm :\longmapsto\:loga = log1$

$\bf\implies \:a = 1$

• Hence, Option (c) is correct

Additional Information :-

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

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

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

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

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