Find the values of a,b,c such that lim_"x→0"ae^(x)-bcosx+ce^(-x)/xsinx=3/2?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

3
Feb 21, 2018

Answer:

#a=c = 3/4, b= 3/2#

Explanation:

#lim_"x→0"{ae^(x)-bcosx+ce^(-x)}/{xsinx}=3/2#

Now, the denominator of the ratio vanishes as #x to 0#. So, for the limit to exist , the numerator must vanish as well.

Thus

#a-b+c = 0#

If this is satisfied, the ratio is of the form #0/0# as #x to 0#, so we could use L'Hospital's rule.

#lim_"x→0"{ae^(x)-bcosx+ce^(-x)}/{xsinx}= lim_{x to 0} {d/dx(ae^(x)-bcosx+ce^(-x))}/{d/dx(s sinx)} = lim_{x to 0}{ae^(x)+b sin x-ce^(-x)}/{sin x+x cos x}#

Once again, the denominator vashes as #x to 0#, so we must have

#a-c = 0#

We apply L'Hospital's rule once more to get

# lim_{x to 0}{ae^(x)+b sinx-ce^(-x)}/{sin x+x cos x} = lim_{x to 0}{d/dx(ae^(x)+b sinx-ce^(-x))}/{d/dx(sin x+x cos x)} = lim_{x to 0}{ae^(x)+b cos x+ce^(-x)}/{2cos x-x sin x} = {a+b+c}/2#

Thus
#a+b+c = 3#

Solving these conditions, we easily get
#a=c = 3/4, b= 3/2#

Was this helpful? Let the contributor know!
1500
Trending questions
Impact of this question
142 views around the world
You can reuse this answer
Creative Commons License