Question

Find the limit as x goes to infinity of
(e^(3x)-e^(-3x))/(e^(3x)+e^(-3x)) or show that it does not exist.

Answer 1

L'hospital's rule

Answer 2

Correct. L'hopital's rule...because as it sits... it = infinity/infinity. Do you know L'Hopital's rule?

Answer 3

it was not covered in class ye,. so maybe it should be limit does not exist?

Answer 4

\[\begin{array}{l} \lim_{x\to \infty } \, \frac{e^{3 x}-\frac{1}{e^{3 x}}}{e^{3 x}+\frac{1}{e^{3 x}}} \\ \text{simplify }\frac{e^{3 x}-\frac{1}{e^{3 x}}}{e^{3 x}+\frac{1}{e^{3 x}}}\text{ assuming }x>0\text{ giving }\frac{e^{6 x}-1}{e^{6 x}+1} \\ \text{= }\lim_{x\to \infty } \, \frac{e^{6 x}-1}{e^{6 x}+1} \\ \text{the limit of a sum is the sum of the other limit} \\ \text{= }\lim_{x\to \infty } \, \frac{e^{6 x}}{1+e^{6 x}}-\lim_{x\to \infty } \, \frac{1}{1+e^{6 x}} \\ the ~limit~ of ~a~ quotient~ is~ the~ quotient ~of ~the~ limits \\ \text{= }\lim_{x\to \infty } \, \frac{e^{6 x}}{1+e^{6 x}}-\frac{1}{\lim_{x\to \infty } \, \left(1+e^{6 x}\right)} \\ \text{Indeterminate form of type $\infty $/$\infty $. Using L'Hospital's rule you'll get} \, \\ \text{= }1 \\\end{array}\]

Answer 5

ok. i can follow the math on that. Teacher has not covered infinty/infinity yet though. thx