Question

If limit approaches infinity, how do you solve with e^x?

Answer 1

We have to solve the limit:

`\lim _(x\to\infty)(e^(7x)+8e^(-3x))/(e^(-3x)+7e^(7x))`

We can divide both numerator and denominator by the biggest exponent, in order to have no term undefined:

`\begin{gathered} \lim _(x\to\infty)(e^(7x)+8e^(-3x))/(e^(-3x)+7e^(7x)) \\ \lim _(x\to\infty)((e^(7x))/(e^(7x))+(8e^(-3x))/(e^(7x)))/((e^(-3x))/(e^(7x))+(7e^(7x))/(e^(7x))) \\ \lim _(x\to\infty)(1+8e^(-10x))/(e^(-10x)+7)=(1+0)/(0+7)=(1)/(7) \end{gathered}`

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