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I came across this particular problem in my coursework, and instead of following the approach given in the answers, I thought of doing it by splitting it using properties of limits and then solving. The question is:

If $\lim_{x \to 0} \dfrac{sin3x}{x^3}+\dfrac{a}{x^2}+b $ exists, and is equal to 0 then find value of a and b

What I did:

$$\lim_{x \to 0} \dfrac{sin3x}{x^3}+\dfrac{a}{x^2}+b = 0$$ $$\lim_{x \to 0} \dfrac{3}{x^2}\dfrac{sin3x}{3x} + \dfrac{a}{x^2} +b = 0$$

As, $\lim_{x \to 0} \dfrac{sinx}{x}=1$, we can reduce it to:

$$\lim_{x \to 0} \dfrac{3}{x^2} + \dfrac{a}{x^2} +b = 0$$

Clearly, for the limit to exist, we will need to eliminate the $x^2$ in the denominator, so in order for that to happen, a must be -3. As a result we get:

$$\lim_{x\to 0}\dfrac{3-3}{x^2} +b = 0$$ Which is just: $$\lim_{x\to 0}b = 0$$

Which must mean b is equal to zero, right? But the solutions given in the above given link, and the graph of the above equation say otherwise. I understand the already-given solutions involving trigonometric substitution, but I can't seem to put my finger on why this particular method fails- is there any conceptual error I made here?

Kritin Thakur
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1 Answers1

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The error lies in reducing$$\lim_{x\to0}\frac3{x^2}\cdot\frac{\sin(3x)}{3x}\quad\text{to}\quad\lim_{x\to0}\frac3{x^2}$$just because $\lim_{x\to0}\frac{\sin(3x)}{3x}=1$. When you are dealing with an expression in which $x$ appears several times, you cannot take some of those $x$'s to $0$, while you leave the remaining $x$'s untouched. Otherwise, you can get things such as$$1=\lim_{x\to0}1=\lim_{x\to0}\frac xx=\lim_{x\to0}\frac0x=0.$$

José Carlos Santos
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  • So does that mean whenever dealing with powers of a variable in the limit, we must reduce the entire term to zero at once? I thought we could split the terms as $\lim_{x \to a} f(x) g(x)$ into $\lim_{x \to a} f(x)$ $\lim_{x \to a} g(x)$, and then proceed, or is that not valid? – Kritin Thakur Nov 09 '21 at 11:21
  • Yes, if both limits $\lim_{x\to0}f(x)$ and $\lim_{x\to0}g(x)$ exist, then$$\lim_{x\to0}\bigl(f(x)g(x)\bigr)=\lim_{x\to0}f(x)\lim_{x\to0}g(x).$$What you cannot do is this: if $\lim_{x\to0}f(x)=l$, then to act as if $f(x)g(x)$ is the same thing as $l\times g(x)$. – José Carlos Santos Nov 09 '21 at 11:25