The value of limx- aax-xa-xx-aa-1- then a equals

The correct option is A 1 lim_x→ aa^x x^ax^x a^a= 1 Using L hospital's rule, ⇒lim_x→ aa^xlog a ax^a 1x^x(1+log x)= 1 ⇒a^alog a a.a^a 1a^a(1+log a) ...

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Standard XII

Mathematics

Logarithmic Inequalities

The value of ...

Question

The value of $lim x \to a \frac{a^{x} - x^{a}}{x^{x} - a^{a}} = - 1$ , then $a$ equals

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Solution

The correct option is A 1
$lim x \to a \frac{a^{x} - x^{a}}{x^{x} - a^{a}} = - 1$
Using L-hospital's rule,
$\Rightarrow {lim}_{x \to a} \frac{a^{x} log a - a x^{a - 1}}{x^{x} (1 + log x)} = - 1$
$\Rightarrow \frac{a^{a} log a - a . a^{a - 1}}{a^{a} (1 + log a)} = - 1$
$\Rightarrow \frac{log a - 1}{log a + 1} = - 1$
$\Rightarrow log a = 0 \Rightarrow a = 1$

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The value of limx→aax−xaxx−aa=−1, then a equals

The correct option is A 1limx→aax−xaxx−aa=−1Using L-hospital's rule,⇒limx→aaxloga−axa−1xx(1+logx)=−1⇒aaloga−a.aa−1aa(1+loga)=−1⇒loga−1loga+1=−1⇒loga=0⇒a=1

The value of $lim x \to a \frac{a^{x} - x^{a}}{x^{x} - a^{a}} = - 1$ , then $a$ equals