Limit exercise asked in the Brillant website and solved without using L'Hopital rule or Series expansion

Exercise:

Find \[\large{\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\sin x-{{\left( \sin x \right)}^{\sin x}}}{1-\sin x+\ln \left( \sin x \right)}}\]

Solution: Let $u=\sin x$ as $x\to \frac{\pi }{2}\,\,,\,\,u\to 1$

So $\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\sin x-{{\left( \sin x \right)}^{\sin x}}}{1-\sin x+\ln \left( \sin x \right)}=\underset{u\to 1}{\mathop{\lim }}\,\frac{u-{{u}^{u}}}{1-u+\ln u}=\frac{0}{0}\,\,\,\,ind\,\,form$

           $=\underset{u\to 1}{\mathop{\lim }}\,\frac{u-{{u}^{u}}}{1-u+\ln u}=\underset{u\to 1}{\mathop{\lim }}\,\frac{\frac{u-{{u}^{u}}}{{{\left( u-1 \right)}^{2}}}}{\frac{1-u+\ln u}{{{\left( u-1 \right)}^{2}}}}=\frac{\underset{u\to 1}{\mathop{\lim }}\,\frac{u-{{u}^{u}}}{{{\left( u-1 \right)}^{2}}}}{\underset{u\to 1}{\mathop{\lim }}\,\frac{1-u+\ln u}{{{\left( u-1 \right)}^{2}}}}=\frac{{{\ell }_{1}}}{{{\ell }_{2}}}$

Where ${{\ell }_{1}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{u-{{u}^{u}}}{{{\left( u-1 \right)}^{2}}}\,\,\,\,\,\,\,\,\And \,\,\,\,\,\,{{\ell }_{2}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{1-u+\ln u}{{{\left( u-1 \right)}^{2}}}$

Let $t=u-1\Leftrightarrow u=t+1$ as $u\to 1\,\,,\,\,t\to 0$

So ${{\ell }_{1}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{u-{{u}^{u}}}{{{\left( u-1 \right)}^{2}}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{u\left( 1-{{u}^{u-1}} \right)}{{{\left( u-1 \right)}^{2}}}=\underset{t\to 0}{\mathop{\lim }}\,\frac{\left( t+1 \right)\left( 1-{{\left( t+1 \right)}^{t}} \right)}{{{t}^{2}}}=\underset{t\to 0}{\mathop{\lim }}\,\frac{1-{{\left( t+1 \right)}^{t}}}{{{t}^{2}}}$

       $=\underset{t\to 0}{\mathop{\lim }}\,\frac{1-{{\left( t+1 \right)}^{t}}}{{{t}^{2}}}\times \frac{t\ln \left( t+1 \right)}{t\ln \left( t+1 \right)}=\underset{t\to 0}{\mathop{\lim }}\,\frac{\ln \left( t+1 \right)}{t}\times \underset{t\to 0}{\mathop{\lim }}\,\frac{t\left( 1-{{\left( t+1 \right)}^{t}} \right)}{{{t}^{2}}\ln \left( t+1 \right)}$

But \(\underset{t\to 0}{\mathop{\lim }}\,\frac{\ln \left( t+1 \right)}{t}=\underset{t\to 0}{\mathop{\lim }}\,\frac{\ln \left( t+1 \right)-\ln \left( 1 \right)}{t-0}=\frac{d}{dt}{{\left( \ln \left( t+1 \right) \right)}_{t=0}}={{\left. \frac{1}{t+1} \right|}_{t=0}}=1\)

So ${{\ell }_{1}}=\underset{t\to 0}{\mathop{\lim }}\,\frac{\ln \left( t+1 \right)}{t}\times \underset{t\to 0}{\mathop{\lim }}\,\frac{1-{{\left( t+1 \right)}^{t}}}{t\ln \left( t+1 \right)}=\underset{t\to 0}{\mathop{\lim }}\,\frac{1-{{\left( t+1 \right)}^{t}}}{t\ln \left( t+1 \right)}$

Let $w=t\ln \left( t+1 \right)\Leftrightarrow {{e}^{w}}={{\left( t+1 \right)}^{t}}$ as $t\to 0,w\to 0$

So ${{\ell }_{1}}=\underset{t\to 0}{\mathop{\lim }}\,\frac{1-{{\left( t+1 \right)}^{t}}}{t\ln \left( t+1 \right)}=\underset{w\to 0}{\mathop{\lim }}\,\frac{1-{{e}^{w}}}{w}=-\underset{w\to 0}{\mathop{\lim }}\,\frac{{{e}^{w}}-1}{w}=-\underset{w\to 0}{\mathop{\lim }}\,\frac{{{e}^{w}}-{{e}^{0}}}{w-0}=-\frac{d}{dw}{{\left( {{e}^{w}} \right)}_{w=0}}=-1$

Thus ${{\ell }_{1}}=-1\times 1=-1$

Now Let’s compute ${{\ell }_{2}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{1-u+\ln u}{{{\left( u-1 \right)}^{2}}}$

Let $u=\frac{1}{t}$ as $u\to 1\,\,,t\to 1$

So ${{\ell }_{2}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{1-u+\ln u}{{{\left( u-1 \right)}^{2}}}=\underset{t\to 1}{\mathop{\lim }}\,\frac{1-\frac{1}{t}+\ln \frac{1}{t}}{{{\left( \frac{1}{t}-1 \right)}^{2}}}=\underset{t\to 1}{\mathop{\lim }}\,\frac{1-\frac{1}{t}-\ln t}{\frac{{{\left( 1-t \right)}^{2}}}{{{t}^{2}}}}$

         $=\underset{t\to 1}{\mathop{\lim }}\,\frac{{{t}^{2}}\left( 1-\frac{1}{t}-\ln t \right)}{{{\left( t-1 \right)}^{2}}}=\underset{t\to 1}{\mathop{\lim }}\,{{t}^{2}}\times \underset{t\to 1}{\mathop{\lim }}\,\frac{1-\frac{1}{t}-\ln t}{{{\left( t-1 \right)}^{2}}}=\underset{t\to 1}{\mathop{\lim }}\,\frac{1-\frac{1}{t}-\ln t}{{{\left( t-1 \right)}^{2}}}$

So $2{{\ell }_{2}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{1-u+\ln u+1-\frac{1}{u}-\ln u}{{{\left( u-1 \right)}^{2}}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{1-u+1-\frac{1}{u}}{{{\left( u-1 \right)}^{2}}}$

          $=\underset{u\to 1}{\mathop{\lim }}\,\frac{\frac{-{{u}^{2}}+2u-1}{u}}{{{\left( u-1 \right)}^{2}}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{-\left( {{u}^{2}}-2u+1 \right)}{{{\left( u-1 \right)}^{2}}}=\underset{u\to 1}{\mathop{\lim }}\,\frac{-{{\left( u-1 \right)}^{2}}}{{{\left( u-1 \right)}^{2}}}=-1$

Thus $\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\sin x-{{\left( \sin x \right)}^{\sin x}}}{1-\sin x+\ln \left( \sin x \right)}=\frac{{{\ell }_{1}}}{{{\ell }_{2}}}=\frac{-1}{\frac{-1}{2}}=2$





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The idea of solution Credit to Amer Khamiseh