Evaluating limit with 3 unknown parameters
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Dr.Peterson
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My first thought would be that the numerator has to go to zero, since the denominator does; that will give you one equation in a, b, and c.
Then you might try L'Hopital's rule, if you know that.
If you can show your work as far as you can get with these thoughts, and also tell us what you have learned that might be applicable, we can consider the next step. (I haven't looked that far yet.)
HallsofIvy
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As Dr. Peterson said, since the denominator goes to 0, in order that the limit exist, the numerator must also go to 0 and that makes it a candidate for "L'Hopital's rule". The derivative of the numerator, with respect to x, is \(\displaystyle -b sin(x)+ 4ae^{4x}- ce^{-x}\) and the derivative of the denominator, with respect to x, is -sin(2x)-2x cos(x). The denominator goes to 0 so, in order that the limit exist, the numerator must also. That gives a second equation for a, b, and c and allows us to use "L'Hopital" a second time. Differentiating the numerator again, we have \(\displaystyle -b cos(x)+ 16ae^{4x}+ ce^{-x}\). Differentiating the denominator again, \(\displaystyle -4 cos(3x)+ 2x sin(x)\).
Finally, as x goes to 0, the denominator goes to -4, not 0 and the numerator goes to -b+ 16a+ c so we must have \(\displaystyle -\frac{-b+ 16a+ c}{4}= 10\).
We have three equations to solve for a, b, and c.
Finally, as x goes to 0, the denominator goes to -4, not 0 and the numerator goes to -b+ 16a+ c so we must have \(\displaystyle -\frac{-b+ 16a+ c}{4}= 10\).
We have three equations to solve for a, b, and c.