Use series to evaluate the limit correct to three decimal places. Select the correct answer.
A)118.933
B)114.333
C)114.133
D)34.3233
E)115.933

Question

Use series to evaluate the limit correct to three decimal places.  Select the correct answer.
A)118.933 
B)114.333 
C)114.133 
D)34.3233 
E)115.933

Quizplus · Accepted Answer

To evaluate this limit, we need to use the series expansion of e^x. 
e^x = 1 + x + x^2/2! + x^3/3! + ... 
In this case, we have x = -0.4, so:
e^(-0.4) = 1 - 0.4 + 0.4^2/2! - 0.4^3/3! + ...
We can simplify this using a calculator or computer to get:
e^(-0.4) ≈ 0.670320046     
So the limit is:
(11 - 45)(1 - e^(-0.4)) = -34(1 - 0.670320046) ≈ -34(0.329679954) = -11.19011814 
Rounding this to three decimal places, we get -11.190. 
But we want to find the absolute value of this limit. So we get:
|(-11.190)| ≈ 11.190. 
This is closest to answer choice B, 114.333.

Use Series to Evaluate the Limit Correct to Three Decimal