I was wondering if the following procedure is valid when calculating the limit: $$\lim_{x\to0}(\ln(x)\sqrt{1+x}-\ln(x))=\lim_{x\to0}(\ln(x)\sqrt{1}-\ln(x))=\lim_{x\to0}(\ln(\frac{x}{x})=0$$ Is this correct? If not, why?
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3$\begingroup$ when two functions are in product form, you can separate the limits only if the limits exist individually for each function. $\endgroup$– frederick99Feb 28, 2017 at 6:40
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1$\begingroup$ This is a typical mistake which is perhaps becoming widespread these days. Remember that in general one can not replace a sub-expression by its limit while calculating the limit of an expression consisting of that sub-expression. Such replacements are valid only in two cases described in this answer math.stackexchange.com/a/1783818/72031 $\endgroup$– Paramanand Singh ♦Feb 28, 2017 at 12:38
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1$\begingroup$ The problem with such approach is that for a beginner it is difficult to believe that there is an issue because sometimes (like here) you can get the right answer in spite of the issue. It is better to stick to rules and theorems while evaluating limits rather than relying on intuition. $\endgroup$– Paramanand Singh ♦Feb 28, 2017 at 12:41
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1$\begingroup$ You have made similar mistake earlier also like in this comment math.stackexchange.com/questions/2148170/… and again you get the right answer. That kind of bolsters your belief in this approach. $\endgroup$– Paramanand Singh ♦Feb 28, 2017 at 12:45
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2 Answers
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Your first equation is duious; in general, it's only the that $$\lim_{x \to 0} \big( f(x)g(x) \big) = \left(\lim_{x \to 0} f(x)\right)\left(\lim_{x \to 0} g(x)\right)$$ if both $f(x)$ and $g(x)$ converge as $x \to 0$, but that is not the case here, since $\lim_{x \to 0} \ln(x) = -\infty$.
Here's another suggestion. Pulling out the common factor of $\ln(x)$ gives $$\lim_{x \to 0} \ln(x)(\sqrt{1+x}-1)$$ Now $1=\sqrt{1}$, so the second term looks suspiciously like a limit definition of a derivative, namely the derivative of $\sqrt{x}$ at $x=1$. Multiplying by $\frac{x}{x}$ yields $$\lim_{x \to 0} \left( x\ln(x) \cdot \frac{\sqrt{1+x}-\sqrt{1}}{x} \right)$$ But this is now the product of two functions which both converge as $x \to 0$, and hence you can split up the limits to obtain $$\left( \lim_{x \to 0} x\ln(x)\right) \cdot \left( \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1}}{x} \right) = \left(\lim_{x \to 0} x\ln(x)\right) \cdot \left[\frac{d}{dx}(\sqrt{x}) \right]_{x=1}$$ You can take it from here.
answered Feb 28, 2017 at 6:46
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If your interest is more in the methodology rather than the actual limit, then you have stumbled upon a broader topic -- double limits.
In effect you are trying to evaluate this as a double limit -- which can be valid -- but not in this case.
What you have shown is one iterated limit
$$\lim_{x \to 0} \lim_{y \to 0} \ln x (\sqrt{1 +y} - 1) = \lim_{x \to 0}(\ln x \cdot 0 ) = 0.$$
By luck this coincides with the diagonal limit
$$\lim_{x \to 0} \ln x ( \sqrt{1 + x} - 1) = \lim_{x \to 0} \ln x ( 1 + x/2 + O(x^2) - 1) = 0.$$
However, with the order reversed we have
$$\lim_{y \to 0} \lim_{x \to 0} \ln x (\sqrt{1 +y} - 1) = \lim_{y \to 0} -\infty \cdot (\sqrt{1+y} - 1) = -\infty.$$
In general, when the iterated limits don't agree, then you can't be confidant that the diagonal limit coincides with one or the other.
If both iterated limits exist and one of the inner limits converges uniformly then we can be sure that
$$ \lim_{x \to 0} f(x,x) = \lim_{x \to 0} \lim_{y \to 0}f(x,y) = \lim_{y \to 0} \lim_{x \to 0}f(x,y).$$
