Question

lim x trend b (x^b-b^x)/( x^x-b^b)​

Answer 1

Answer:

x/b (b)^(x-b)

Step-by-step explanation:

lt x trend to b (x^n-a^n)/(x-a)= n.a^(n-1)

Answer 2

The value of  $\lim_{x \to b} \frac{x^{b}-b^{x} }{x^{x} -b^{b} }$   is   $\frac{(1-\log b)}{( \log b-1)}$ .

Step-by-step explanation:

Given:

$\lim_{x \to b} \frac{x^{b}-b^{x} }{x^{x} -b^{b} }$

To Find:

The value of  $\lim_{x \to b} \frac{x^{b}-b^{x} }{x^{x} -b^{b} }$ .

Formula Used:

As per  L' Hospital rule-

If $\lim_{x \to \ a} \frac{f(x)}{g(x)} =\frac{0}{0} \ or\ \frac{\infty}{\infty}$,   then $=\lim_{x \to \ a} \frac{f'(x)}{g'(x)}$ .

$\frac{d}{dx} x^{n} = nx^{n-1}$

$\frac{d}{dx} a^{x} = a^{x} \log a$

Solution:

As given - $\lim_{x \to b} \frac{x^{b}-b^{x} }{x^{x} -b^{b} }$ .

$\lim_{x \to b} \frac{x^{b}-b^{x} }{x^{x} -b^{b} } = \frac{b^{b}-b^{b} }{b^{b} -b^{b} } =\frac{0}{0}$

Then applying the L' Hospital rule.

      $= \lim_{x \to b} \ \frac{\frac{d}{dx}(x^{b}-b^{x}) }{\frac{d }{dx}(x^{x}-b^{b} ) }$

      $= \lim_{x \to b} \ \frac{\frac{d}{dx}(x^{b})-\frac{d}{dx} (b^{x}) }{\frac{d }{dx}(x^{x})- \frac{d}{dx} (b^{b} ) }$                      

      $= \lim_{x \to b} \frac{( bx^{b-1} - b^{x}\log b )}{x^{x}( logx+1)-0 }$                            $Let \y=x^{x} \\\log y= x \log x\\On\ differentiating \ w.r.t. x, we\ get.\\\frac{1}{y} \frac{dy}{dx} =\frac{x}{x} + \log x\\ \frac{dy}{dx} =y( 1+ \log x)\\ \frac{d(x^{x} )}{dx} =x^{x} ( 1+ \log x)$        

      $= \frac{( b\ b^{b-1} - b^{b}\log b )}{b^{b}( \log b+1)-0 }$

      $= \frac{( \ b^{b} - b^{b}\log b )}{b^{b}( \log b+1) }$

      $= \frac{ \ b^{b}(1 - \log b )}{b^{b}( \log b+1) }$

      $= \frac{ \ (1 - \log b )}{( \log b+1) }$

Thus,the value of  $\lim_{x \to b} \frac{x^{b}-b^{x} }{x^{x} -b^{b} }$   is   $\frac{(1-\log b)}{( \log b-1)}$ .