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Show that $$|x-1|\sin{3\over(x-1)} + \ln(x)= O(x-1)$$ as $x$ approaches $1$.

I have a hard time understanding Big-O and Small-o notations in Calculus. I have tried to prove this by using limits, considering two cases with module and I've got $$M \leq 2$$

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Hint: What needs to be shown is that the quotient of the left hand side expression and $x - 1$ can be bounded as $x \to 1$.

By L'Hospital's rule,

$$\lim_{x \to 1} \frac{\ln x}{x - 1} = \lim_{x \to 1} \frac{1}{x} = 1.$$ and notice that $$\left|\sin(1/(x - 1))\right| \leq 1.$$

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    $\begingroup$ You don't need L'Hospital's rule for the limit: it's the very definition of the derivative of $\ln x$ at $x=1$. $\endgroup$
    – Bernard
    Commented Oct 22, 2015 at 19:51

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