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Question:

If $$\lim_{x\to 0}{\left[\frac{f(x)}{\sin(2x)}\right]}=2$$ and $$\lim_{x\to 0}{\left[(\sqrt{x+4}-2)\cdot{g(x)}\right]}=5$$ then find $$\lim_{x\to 0}{[f(x)\cdot{g(x)}]}$$

But, from what I found, $\lim_{x\to 0}{[g(x)]}$ does not exist. So how can I find $\lim_{x\to 0}{[f(x)\cdot{g(x)}]}$?

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Only one theorem applied.

Theorem. If $\lim\limits_{x\rightarrow c}f(x)=L$ and $\lim\limits_{x\rightarrow c}g(x)=M$, then $\lim\limits_{x\rightarrow c}[f(x)g(x)]=LM$

$$ \because \qquad \lim_{x\rightarrow0}\frac{f(x)}{\sin(2x)}=2\quad \text{and}\quad \lim_{x\rightarrow0}\left[(\sqrt{x+4}-2)\cdot g(x)\right]=5 $$

$$ \therefore\qquad \lim_{x\rightarrow0}\left\{\frac{f(x)}{\sin(2x)}\cdot \left[(\sqrt{x+4}-2)\cdot g(x)\right]\right\}=2\times5=10 $$

Rearranging: \begin{align} \frac{f(x)}{\sin(2x)}\cdot \left[(\sqrt{x+4}-2)\cdot g(x)\right] &=\left[f(x)g(x)\right]\cdot\frac{\sqrt{x+4}-2}{\sin(2x)}\\ &=\left[f(x)g(x)\right]\cdot\frac{2x}{\sin(2x)}\cdot \frac{1}{2(\sqrt{x+4}+2)} \end{align}

$$ \Rightarrow\qquad \lim_{x\rightarrow0}\left\{\left[f(x)g(x)\right]\cdot\frac{2x}{\sin(2x)}\cdot \frac{1}{2(\sqrt{x+4}+2)}\right\} =10 $$

$$ \because\qquad \lim_{x\rightarrow0}\frac{\sin(2x)}{2x}=1\quad \text{and} \quad \lim_{x\rightarrow0}\frac{2(\sqrt{x+4}+2)}{1}=8 $$

$$ \therefore\qquad \lim_{x\rightarrow0} \left\{ \left[\left[f(x)g(x)\right]\cdot\frac{\sin(2x)}{2x}\cdot \frac{1}{2(\sqrt{x+4}+2)}\right] \cdot \left[ \frac{2x}{\sin(2x)} \right] \cdot \left[ \frac{2(\sqrt{x+4}+2)}{1} \right] \right\}\\=10\cdot 1\cdot8=80 $$ $$ \Rightarrow\qquad \lim_{x\rightarrow0}\left[f(x)g(x)\right]=80 $$

Tips. Treat the block of functions as one function.

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Hint:

$$ \lim_{x \to 0}\left( \frac{f(x)}{\sin(2x)} \cdot (\sqrt{x+4}-2)g(x) \right) = 2\cdot 5 = 10\\ \lim_{x \to 0} \left( \frac{\sqrt{x+4}-2}{\sin(2x)}\cdot f(x)\cdot g(x)\right) = 10.$$

Compute $\lim_{x\to 0} \frac{\sqrt{x+4}-2}{\sin(2x)}$.

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