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Given function $f:\mathbb R→\mathbb R$ for which $|f(x)-2|≤x^2$.

Find the limits:

  1. $\lim_{x\to0}f(x)$.
  2. $\lim_{x\to0}\frac{f(x)-\sqrt{x+4}}{x}$.

I can solve question (a) very easily (squeeze theorem) and the answer is $\lim_{x \rightarrow 0} f(x)= 2$.

But when I start solving question (b), I'm in a dead end. The book says that the answer of question (b) is $\lim_{x \rightarrow 0 }\frac {f(x)-\sqrt{x+4}}{x} = -\frac{1}{4}$.

Perhaps I'm doing an algebra mistake somewhere because I end up always at $-\frac{1}{0}$ which is not possible. Could someone please give me a hint ?

Thanks

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You can split your limit into two parts:

Part 1: $\frac{f(x)-2}{x}$ which tends to zero.

Part 2: $\frac{2-\sqrt{x+4}}{x}$ which tends to $-1/4$ by L'Hopital's rule.

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  • $\begingroup$ I haven't studied L'hopitals rule yet. I'm suppose to solve it without using L'Hopital rule. Many thanks $\endgroup$ – Skoutas May 3 '16 at 9:53
  • $\begingroup$ Why need L'Hospital for such a simple limit? Just multiply the fraction by $2+\sqrt{x+4}$. $\endgroup$ – user26857 May 3 '16 at 10:48
  • $\begingroup$ Or observe that if $g(x) = \sqrt {4+x},$ then the second limit is $-f'(0) $ by the definition of the derivative. $\endgroup$ – zhw. May 3 '16 at 19:21
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a) $lim f(x)$ for $x→0 =2$ because

$0<=|f(x)-2|<=x^2$, by squeeze theorem,

as $x$ goes to $0,0<=|f(x)-2|<=0. so |f(x)-2|=0$,

hence $lim f(x)$ for $x→0 =2$

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  • $\begingroup$ The OP said clearly that he solved part (a). $\endgroup$ – user26857 May 3 '16 at 10:44

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