Limit Exercise Asked by Mohamad Mathe in the math group

Exercise:

Show that , $\underset{x\to \infty }{\mathop{\lim }}\,\frac{1+{{2}^{x}}}{1+{{3}^{x}}}=0$

Solution: we $\underset{x\to \infty }{\mathop{\lim }}\,\frac{1+{{2}^{x}}}{1+{{3}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{1+{{3}^{x}}}+\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{2}^{x}}}{1+{{3}^{x}}}$

Since $\underset{x\to \infty }{\mathop{\lim }}\,\left( 1+{{3}^{x}} \right)=\infty $ so $\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{1+{{3}^{x}}}=\frac{1}{\infty }=0$

So $\underset{x\to \infty }{\mathop{\lim }}\,\frac{1+{{2}^{x}}}{1+{{3}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{2}^{x}}}{1+{{3}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\frac{{{2}^{x}}}{{{3}^{x}}}}{\frac{1+{{3}^{x}}}{{{3}^{x}}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{\left( \frac{2}{3} \right)}^{x}}}{\frac{1}{{{3}^{x}}}+{{1}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{2}{3} \right)}^{x}}$

Put $y={{\left( \frac{2}{3} \right)}^{x}}\Leftrightarrow y={{e}^{x\ln \left( 2/3 \right)}}\Leftrightarrow y=\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{x\ln \left( 2/3 \right)}}$

As $e$ is continuous function then $y={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,x\ln \left( 2/3 \right)}}={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,-x\ln \left( 3/2 \right)}}={{e}^{-\infty }}=0$

So $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{2}{3} \right)}^{x}}=0$ thus $\underset{x\to \infty }{\mathop{\lim }}\,\frac{1+{{2}^{x}}}{1+{{3}^{x}}}=0$