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# A limit problem

I have a problem in limit that made me curious until now. The problem is about the limit and trigonometry. I don't know how to simplify it.. pls help me to solve it. thanks..:)

Note by Leonardo Chandra
3 years, 5 months ago

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I assume $$c$$ is a finite value

The key here is L'hôpital's Rule (3 times)

Substitution of $$x=1$$ must give an indeterminate form of $$\frac {0}{0}$$ because denominator yields $$0$$

So the numerator is equals to $$0$$ when $$x=1 \Rightarrow a+b+1=0$$

Apply L'hôpital's Rule and chain rule, the limit becomes

$$\frac {4 a x^{3} + 3 b x^{2}} { \pi (x - 1) \cos (\pi x) + \sin (\pi x) }$$

And it must yield an indeterminate form of $$\frac {0}{0}$$ as well because denominator yields $$0$$ at $$x=1$$, thus numerator must yield $$0$$ as well

So, $$4a+3b=0$$, solve the two equations gives $$a = 3$$, $$b=-4$$

Now substitute those two values into the limit and apply L'hôpital's Rule Rule once again

$$\frac {36 x^{2} - 24 x } { \pi [ -(x-1) \pi \sin (\pi x) + \cos (\pi x) ] + \pi \cos (\pi x) }$$

Substitute $$x=1$$ gives $$c = - \frac {6}{\pi}$$

Hence, $$a = 3, b=-4, c = - \frac {6}{\pi}$$

By the way, I'm new to LaTeX, anyone know how to increase size of the mathematical symbols used? The fractions displayed are too small in standard font size. · 3 years, 5 months ago

thanks for the answer, Pi Han..I never consider the using of 0/0 to evaluate the value of numerator, then apply L'hopital more than twice to find the answer. Anyway, thanks for the great answer..I need to learn more again about the L'Hopital rule. :) · 3 years, 5 months ago

yeah agreed......with the answer...:) · 3 years, 5 months ago

from where did u get this problem? · 3 years, 5 months ago

I got this problem from my calculus book. this problem is about limit problems..an introduction to improper form of limit..:) · 3 years, 5 months ago

How about We asume that limit when $$x \rightarrow 1$$ is 0/0. Then, calculate it using L'Hopital Theorem.. Sorry, I haven't tried that before.. · 3 years, 5 months ago

since denominator at x=1 tends to 0 then for the limit to exist ,numerator has to be 0 or else it will form indeterminate form....

0/0 is a basic form in limits....:) · 3 years, 5 months ago

thank you very much for explanation :) · 3 years, 5 months ago