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Find a value for a so that the following function is continuous at x = 0: f(x) ={ (x^2 + x sin(x))/x^2 for x /= 0, a for x = 0}

 Feb 6, 2017
 #1
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Well I can tell you that a is 2, but as for how to prove that I kind of forgot. I can see that it is 2 by looking at the graph, but I don't know how to find it by just manipulating the function. Somehow you need to get x out of the denominator and then you plug in 0 for x and you should get 2.

 Feb 6, 2017
 #2
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Okay I looked in my book and figured out some stuff. What you need to do is take the limit of the function as x approaches 0. That basically just means plug in 0 for x. You end up with 0/0, which is indeterminite. But you can apply L'Hopital's Rule to this because the indeterminite form is either 0/0 or infinity/infinity.

 

L'Hopitals Rule says that the limit as x approaches some number c of the numerator divided by the denominator is equal to the limit as x approaches c of the derivative of the numerator divided by the derivative of the denominator.

This is it:

\(\lim_{x \to c}\frac{f(x)}{g(x)}=\lim_{x \to c}\frac{f'(x)}{g'(x)}\)

 

Soooo in your case

f(x)=x2 + xsinx

g(x) = x2

c=0

 

then you need to find f'(x) and g'(x).

 

f'(x)=2x+sinx+xcosx

g'(x)= 2x

 

then just plug it in plug it in

 

\(\lim_{x \to 0}\frac{2x+sinx+xcosx}{2x}\)

 

If you apply L'Hopitals Rule again you get

 

\(\lim_{x \to 0} \frac{2+cosx-xsinx+cosx}{2}\)

 

then plug in 0 for x and you get

\(\frac{2+1-0+1}{2}\)

which is 4/2

 

which is.... dun dun dun..... 2!!!! :)))

 Feb 6, 2017

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