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Question I couldn't solve this question so I looked for hints. One method of solving was to use the Taylor Series expansion of each of the functions. It was a bit long. So another solution used the L'Hospitals Rule instead.

$$\lim_{x\to 0}{\cfrac{\alpha e^x+\beta e^{-x}+\gamma\sin(x)}{x\sin^2x}}$$

$$= \lim_{x\to 0}{\cfrac{\alpha e^x+\beta e^{-x}+\gamma\sin(x)}{x^3}}$$

For limit to be finite limit of numerator should be zero. Therefore $$\alpha +\beta =0.$$ But limit of $\sin x$ as $x$ tends to $0$ is taken as $x$ in most cases so the correct equation should be $$\alpha +\beta +\gamma x=0$$and further equations are obtained similarly.

Can the limit of $\sin x$ as $x$ tends to $0$ be taken as either $x$ or $0$ as per convenience?
Is using L'Hopitals Rule a correct method of solving this question?
OR Is Taylor series the more appropriate method?

Aurelius
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    What's so bad about using Taylor series? You only need the first few (hint: denominator) terms to decide how $\alpha,\beta,\gamma$ relate to get the right amount of cancellation as $x\to0$. – user170231 Mar 05 '24 at 22:53
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    I applied L'Hôpital's rule 3 times, then found a unique answer among the 4 options. Without rewriting $\sin x$ as $x$. Without cancelling factors in the numerator and denominator. – peterwhy Mar 05 '24 at 23:29
  • Related and may help address the first part of your question – user170231 Mar 06 '24 at 06:59
  • @user170231 Nothing bad. I felt it was a long. Also sometimes I forget the series for a function then I have to derive it using $$\Sigma{\cfrac{f^n(a)(x-a)^n}{n!}}$$ – Aurelius Mar 06 '24 at 13:41
  • @peterwhy Using Taylor Series or using $sinx=0$ condition for limit to be finite is that it is an indeterminate form so limit of numerator should be zero, then equation is $$\alpha+\beta=0$$ and then L'Hopital. If $sinx=x$, then the equation is $$\alpha+\beta+\gamma x=0$$ and then L'Hopital. If the equations are different in for different values of limit of $\sin x$. How can they be solved to give same values of $\alpha,\beta,\gamma$ – Aurelius Mar 06 '24 at 13:50
  • P.S. Both the solutions I have find the values of $\alpha,\beta,\gamma $ and then substitute them in options – Aurelius Mar 06 '24 at 13:52
  • @Aurelius Fortunately here you only really need to remember one series for $e^x$, then you can recover $e^{-x}$ by replacing $x\to -x$, and $\sin x$ by using the Euler identity$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ – user170231 Mar 06 '24 at 16:57
  • You can't replace the $\sin x$ by $x$ in numerator, but you can do that in the denominator and one should understand how this works. Further is is best to split the numerator into two parts by adding and subtracting $\gamma x$ and thereby write the expression under limit as a sum of two expressions with one of them having a limit. – Paramanand Singh Mar 07 '24 at 02:33

1 Answers1

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Use B.Taylor is also short!

$e^x = 1 + x/1! + x^2/2! + x^3/3! + ...$

$e^{-x} = 1 - x/1! + x^2/2! - x^3/3! + ...$

$\sin(x)= x - x^3/3! + x^5/5! + ...$

and in the denominator $\sim x^3$.

Hence we have $\alpha+\beta = 0$.

In the numerator: $2\alpha(x+x^3/3!+...) + \gamma(x-x^3/3!+...)$

$\implies 2\alpha + \gamma = 0$ and $2\alpha - \gamma = 4$

$\implies \alpha = 1, \beta = -1, \gamma = -2$.

Therefore, (c) is wrong!

Aig
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looyer
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