1
$\begingroup$

Let $h$ and $j$ be two functions of domain $\mathbb{R}$. It is known that $\lim_{x\rightarrow1}h(x)=0$, $\lim_{x\rightarrow1}j(x)=b$ and $\lim_{x\rightarrow1}\frac{h(x)}{j(x)}=2$.

Only one of the following values can be $b$. Which one?

a)0 $\\$

b)1

c)2

d)$+\infty$

$\\$

I know that $\lim_{x\rightarrow c}\frac{f(x)}{g(x)} = \frac{\lim_{x\rightarrow c} f(x)}{\lim_{x\rightarrow c} g(x)}$

So I get:

$$\lim_{x\rightarrow 1}\frac{h(x)}{j(x)} = \frac{\lim_{x\rightarrow 1} h(x)}{\lim_{x\rightarrow 1} j(x)} = \frac{0}{b}$$

But how does this equal two? My book says the solution is zero. Is this because L'Hôpital's rule is applied? Because if the answer is zero, then $$\lim_{x\rightarrow 1}\frac{h(x)}{j(x)} = \frac{0}{0}$$ which seems a good candidate for L'Hôpital's rule.

In short, my question is why is the right answer zero?

  • 0
    Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/53453/discussion-on-question-by-silenceonthewire-it-is-known-that-lim-x-rightarrow1). – 2017-02-11

1 Answers 1

3

Since $\lim_{x\to 1}h(x)/j(x)=2$, it follows that both $h(x), j(x) $ are non-zero in some deleted neighborhood of $x=1$. Now use algebra of limits to get $$\lim_{x\to 1}j(x)= \lim_{x\to 1}\frac{h(x)}{h(x)/j(x)}=\frac{0}{2}=0$$ Observe that the above algebraic manipulation works only because $h, j$ are non-zero as $x\to 1$. Also note that the above proves that the limit of $j(x) $ exists and is $0$ and hence $b=0$.

It is better to use algebra of limits rather than trying to use intuition and think about "what if $b\neq 0$". Such arguments while correct are indirect and use the assumption that limit of $j$ exists (this is given in question BTW). If this assumption is not given then such arguments can not be made. Thus to sum up, rules of algebra of limits are much more useful and powerful than most beginners think they are.