Question

Please help me find the following limit, click to see...

Answer 1

\[\lim_{b \rightarrow \infty}e ^{-sb+b+7}\]

Answer 2

e^7 is a constant number
we can write the problem as
\[ e^7\lim_{b \rightarrow \infty}e ^{-sb+b} \]
rewrite the exponent by factoring out the b
-sb+b= (-s+1)b = b(1-s)
\[ e^7\lim_{b \rightarrow \infty}e ^{b(1-s)} \]
1-s is also a constant, so write the problem as
\[ e^7e^{(1-s)}\lim_{b \rightarrow \infty}e ^{b} \]

Answer 3

ok thanks

Answer 4

unless s is a complex variable. In which case you need |s|<1

Answer 5

well this is part of a laplace transform

Answer 6

oh, so s is a complex variable.

Answer 7

i'm trying to find the laplace transform of
\[f(t)=e ^{t+7}\]

Answer 8

I know a few things. e^7 is a constant to you can pull it out
L(f(t)) = e^7 L(e^t)
from tables we know L(e^t) is 1/(s-1)

I'll look at doing the integral

Answer 9

phi could you help me with my question after your done helping Babyslapmafro

Answer 10

see http://www.khanacademy.org/math/differential-equations/laplace-transform/laplace-transform-tutorial/v/laplace-transform-2

Answer 11

\[e^7 \int\limits_{0}^{\infty}e^t e^{-st} dt= e^7 \int\limits_{0}^{\infty} e^{(1-s)t} dt\]
ignoring the e^7 for the moment
\[\frac{ 1 }{ 1-s }\left(\lim_{b \rightarrow \infty}\left( e^{(1-s)b}- e^{(1-s)0} \right) \right)= \frac{ 1 }{ 1-s }\left(\lim_{b \rightarrow \infty}e^{(1-s)b}-1\right)\]

Answer 12

finally limit of e^(1-s)b as b-> infinity is 0 as long as 1-s<0 or |1-s|<1
you get
\[ \frac{1}{1-s}(0-1) = - \frac{1}{1-s} = \frac{1}{s-1} \]

Answer 13

multiply by e^7 for the final answer
\[ \frac{e^7}{s-1} \]