We have to solve the limit:
[tex]\lim _{x\to\infty}\frac{e^{7x}+8e^{-3x}}{e^{-3x}+7e^{7x}}[/tex]
We can divide both numerator and denominator by the biggest exponent, in order to have no term undefined:
[tex]\begin{gathered} \lim _{x\to\infty}\frac{e^{7x}+8e^{-3x}}{e^{-3x}+7e^{7x}} \\ \lim _{x\to\infty}\frac{\frac{e^{7x}}{e^{7x}}+\frac{8e^{-3x}}{e^{7x}}}{\frac{e^{-3x}}{e^{7x}}+\frac{7e^{7x}}{e^{7x}}} \\ \lim _{x\to\infty}\frac{1+8e^{-10x}}{e^{-10x}+7}=\frac{1+0}{0+7}=\frac{1}{7} \end{gathered}[/tex]
The value of e^-10x and its multiples, when x tends to infinity, tends to 0, so the value of the limit is 1/7.
Answer: 1/7
