I'm having a bit of a problem taking the limit of functions involving $\sin(\dfrac1x)$.
Mainly, I don't know whether or not L'Hopital's rule is required. Here's the particular problem:
$$\lim\limits_{x\to \infty} 4x\sin (\frac1x)= ?$$
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3Set $1/x=h,h\to0^+$ – lab bhattacharjee Apr 11 '19 at 02:06
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1write your limit as $$4\lim_{?}\frac{\sin(\frac{1}{x})}{\frac{1}{x}} $$ – Chinnapparaj R Apr 11 '19 at 02:09
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4when I hear "inverse trig functions" I think of $\sin^{-1}(x)$ et al. – J. W. Tanner Apr 11 '19 at 02:11
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I edited the question accordingly – J. W. Tanner Apr 11 '19 at 03:51
2 Answers
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$$ \lim_{x\to \infty}\left[4x\sin\left(\frac1x\right)\right]= 4\lim_{x\to \infty}\frac{ \sin\left(\frac1x\right)}{\frac1x} $$
As $x\to\infty$, $\frac1x \to0$. Substitute $\frac1x$ with a new variable ($\theta=\frac1x$):
$$ 4\lim_{\theta\to 0}\frac{ \sin\theta}{\theta}=4\cdot 1=4. $$
$\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ is a well-known limit and the fact that it's equal to 1 is proven in elementary calculus.
Or you can use L'Hospital's rule (the indeterminate form is $\frac00$):
$$4\lim_{x\to\infty}\frac{\sin\left(\frac1x\right)}{\frac1x}\stackrel{\text{L'H}}{=} 4\lim_{x\to\infty}\frac{\left[\sin\left(\frac1x\right)\right]'}{\left(\frac1x\right)'}=4\lim_{x\to\infty}\frac{\cos{\left(\frac1x\right)}\left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}=\\ 4\lim_{x\to\infty}\cos{\left(\frac1x\right)}=4\cdot\cos{0}=4\cdot 1=4. $$
Michael Rybkin
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Thanks, another thing: could you use l'hopital's for this or is that restricted to rational functions? – alipboy Apr 11 '19 at 02:24
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Thanks so much! This is way more in line with how we're learning it in school – alipboy Apr 11 '19 at 02:35
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To the proposer: Review the necessary and sufficient conditions on the functions in the complete statement of l'Hopital's Rule. Observe that those conditions are very broad and do not exclude any particular type (i.e. trig, exponential,etc.) of functions. – DanielWainfleet Apr 11 '19 at 19:19
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Let $h=1/x$. As $x\to \infty$, $h\to 0$.
$$\lim_{x\to \infty}4x\sin\left(\dfrac{1}{x}\right)=4\cdot\lim_{x\to \infty}\dfrac{\sin\left(1/x\right)}{1/x}\mapsto4\cdot\underbrace{\lim_{h\to 0}\dfrac{\sin h}{h}}_{\to 1}\to 4$$
Paras Khosla
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