Here is my limit to be evaluated
$\lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right)$=?
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Hint : Let $h = \dfrac{1}{x}$ and notice that your limit is the same as $\lim\limits_{h \to 0} (2 + 2 \cdot\dfrac{\sin(4 h) - \sin(0)}{h - 0})$ – Esperluet Dec 06 '15 at 13:56
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What was wrong ill remember it?? – Archis Welankar Dec 06 '15 at 14:18
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Here is a simple step by step approch
$$\begin{align} &\quad \lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right) \\ &= \lim_{x \to \infty} 2 + 8 \frac{x}{4} \sin\left(\frac{4}{x}\right) \\ &= \lim_{x \to \infty} 2 + 8 \lim_{x \to \infty} \frac{\sin\left(\frac{4}{x}\right)}{\frac{4}{x}} \\ &= \lim_{x \to \infty} 2 + 8 \lim_{u \to 0} \frac{\sin(u)}{u} \\ &=2+8(1)=10 \end{align} $$
Hosein Rahnama
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BIG HINT:
$$\lim_{x\to\infty}\left(2+2x\sin\left(\frac{4}{x}\right)\right)=2+\left(\lim_{x\to\infty}2x\sin\left(\frac{4}{x}\right)\right)=2+2\left(\lim_{x\to\infty}x\sin\left(\frac{4}{x}\right)\right)=$$
Let $n=\frac{1}{x}$:
$$2+2\left(\lim_{n\to 0}\frac{1}{n}\cdot\sin\left(4n\right)\right)=2+2\left(\lim_{n\to 0}\frac{\sin\left(4n\right)}{n}\right)=$$ $$2+2\left(\lim_{n\to 0}\frac{\frac{\text{d}}{\text{d}n}\left(\sin\left(4n\right)\right)}{\frac{\text{d}}{\text{d}n}\left(n\right)}\right)=2+2\left(\lim_{n\to 0}\frac{4\cos(4n)}{1}\right)=2+8\lim_{n\to 0}\cos(4n)$$
Jan Eerland
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The limit in question is $2+2\lim_{x\to\infty}x\sin\frac{4}{x}=2+8=10$.
MickG
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Archis Welankar
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For all $n$$$2+2x\sin (\frac nx)=2+2 \frac{\sin (\frac nx)}{\frac 1x}=2+2n \frac{\sin (\frac nx)}{\frac nx}\rightarrow 2+2n$$
Piquito
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