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enter image description here The given function is not defined at $x = 0$. How should the function be defined at $x = 0$ to make it continuous at $x = 0$. Neither the use of expansion of trigonometric functions nor L'Hospital’s rule is allowed.

If these are not allowed how can we solve it

Please anybody can help me in this.

Koolman
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  • The first step is to reduce to one fraction : $$f(x)=\frac{2x+x\cos(x)-3\sin(x)}{x^4\sin(x)}$$ – Peter Dec 22 '16 at 17:43
  • @Peter yeah this is easy , but how to proceed further – Koolman Dec 22 '16 at 17:44
  • @Peter as given in the question without the use of series – Koolman Dec 22 '16 at 17:48
  • and than? i don't see how we can solve this without L'Hospital it is $$\frac{1}{60}$$ – Dr. Sonnhard Graubner Dec 22 '16 at 17:48
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    I just noticed that I imisinterpreted the text. Apparantly, both methods are forbidden. I interpreted Use the expansiion - and L'hospital is not allowed ... – Peter Dec 22 '16 at 17:54
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    @Dr.SonnhardGraubner We would not need L'hospital, if series would be allowed. We could factor out $x^5$ in the numerator and the denominator and would easily get the limit. – Peter Dec 22 '16 at 17:58

2 Answers2

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Let $g(x)=2x+x\cos x-3\sin x.$ Let $g^i$ denote the $i$th derivative of $g$.

We have $g(0)=g^1(0)=g^2(0)=g^3(0)=g^4(0)=0 .$ And $g^5(x)=2\cos x-x\sin x,$ so $\lim_{x\to 0}g^5(x)=2.$

For any $r>0$ take $s>0$ such that $|x|<s\implies 2-r<g^5(x)<2+r.$

Now for $0\leq x<r$ we have $g^4(x)=g^4(x)-g^4(0)=\int_0^xg^5(y)dy,$ so $$(2-r)x\leq g^4(x)\leq(2+r)x .$$ We have $g^3(x)=g^3(x)-g^3(0)=\int_0^x g(y)dy,$ which is bounded above and below by $\int_0^x(2\pm r)y\;dy.$ That is, $$(2-r)x^2/2\leq g^3(x)\leq (2+r)x^2/2.$$ Continuing in this manner for three more steps, we obtain $$x^5(2-r)/120\leq g(x)\leq x^5(2+r)/120$$ for $0\leq x<s.$ Similarly $$x^5(2+r)/120\leq g(x)\leq x^5(2-r)/120$$ for $-s<x<0.$

I assume we do not have to prove that $\lim_{x\to 0}x/\sin x=1.$

For $x\ne 0$ we have $f(x)=(g(x)/x^4)(x/\sin x).$ So we should define $f(0)=0$ to make $f$ continuous at $0.$

DanielWainfleet
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This is too long for a comment, but it is not a complete answer. This shows one type of trick that you can use to make things simpler.

Consider $$ \lim_{x\rightarrow 0}\left(\frac{2+\cos(x)}{x^3\sin(x)}-\frac{3}{x^4}\right)= \lim_{x\rightarrow 0}\left(\frac{-1+\cos(x)}{x^3\sin(x)}+\frac{3}{x^3\sin(x)}-\frac{3}{x^4}\right). $$ Multiplying the numerator and denominator of the first term by the conjugate $1+\cos(x)$, we get that this limit equals $$ \lim_{x\rightarrow 0}\left(\frac{-1+\cos^2(x)}{x^3\sin(x)(1+\cos(x))}+\frac{3}{x^3\sin(x)}-\frac{3}{x^4}\right) $$ Since $-1+\cos^2(x)=-\sin^2(x)$, we have that this simplifies to $$ \lim_{x\rightarrow 0}\left(\frac{-\sin^2(x)}{x^3\sin(x)(1+\cos(x))}+\frac{3}{x^3\sin(x)}-\frac{3}{x^4}\right)=\lim_{x\rightarrow 0}\left(\frac{-\sin(x)}{x^3(1+\cos(x))}+\frac{3}{x^3}\left(\frac{1}{\sin(x)}-\frac{1}{x}\right)\right). $$ What's nice is that the first term is now, essentially, $\frac{-1}{2x^2}$, which is a lower power in the denominator. Other trig identities should allow this to continue to simplify (hopefully).

Michael Burr
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