# Two different answers with different methods in solving limits.

So the question is, $$\lim_{x\to 0}\{\frac{(1+x){\log(1+x)} - x}{x^2}\}$$

Method 1: Applying l'hopital's rule the limit tend to $$\bigl(\frac{1}{2}\bigr)$$.

Method 2: Solving the following way: $$\lim_{x\to 0} \biggl(\frac{(1+x)\log(1+x)}{x^2}\biggr)-\lim_{x\to 0} \biggl(\frac{1}{x}\biggr)$$ $$\lim_{x\to 0} \biggl(\frac{1+x}{x}\biggr)\biggl(\frac{\log(1+x)}{x}\biggr)-\lim_{x\to 0} \biggl(\frac{1}{x}\biggr)$$ $$\lim_{x\to 0} \biggl(\frac{1}{x} + 1\biggr)\biggl(\frac{\log(1+x)}{x}\biggr)-\lim_{x\to 0} \biggl(\frac{1}{x}\biggr)$$ using property: $$\lim_{x\to 0} \Biggl(\frac{\log(1+x)}{x}\Biggr) = 1$$ $$\lim_{x\to 0} \biggl(\frac{1}{x} + 1\biggr)-\lim_{x\to 0} \biggl(\frac{1}{x}\biggr)$$ $$\lim_{x\to 0} \biggl(\frac{1}{x} + 1-\frac{1}{x}\biggr)$$ $$= 1$$ So, why is the limit different for different methods ?

• the answer using l'hopital's rule is actually $1/2$ – tortue Feb 5 '19 at 15:36
• Also, whatever else is at issue in method 2, be careful when you split a limit in 2, if doing so results in terms with $\infty$: for instance, it is clear $\lim_{x\to 0^+}{1\over x} = \infty$. Less clear, but $\lim_{x\to 0^+} {(1 +x )\log( 1 + x)\over x^2} =\infty$, also. So you have [sort of] written down "$\infty -\infty$," which is meaningless as it stands. (On the other hand, "$4-\infty$" is meaningful.) – peter a g Feb 5 '19 at 15:46
• @peterag "$\infty - \infty$" is not meaningless if they are from the same function, in this case, $\frac{1}{x} - \frac{1}{x}$. – Rupanshu Yadav Feb 5 '19 at 15:59
• @RupanshuYadav - $\infty - \infty$ is meaningless on its own. $\lim_{x\to0} \frac1x - \lim_{x\to0} \frac1x$ is not defined, it is not zero. – Glen O Feb 5 '19 at 16:01
• @RupanshuYadav : It is important, conceptually speaking, to bear in mind that a limit, if it exists, is a number ($+\infty$ and $-\infty$ are numbers for this discussion). Now, a number does not 'remember' how it was calculated. Therefore, in calculus, one should think "$\infty - \infty = \cdots$" doesn't scan... It's almost a question of grammar, rather semantics. You cannot recombine the difference into a single expression (the $\cdots$ on the RHS of the $=$) - you're not 'allowed' to know where the $\infty$ came from, once you wrote it down. I hope this is clear! – peter a g Feb 5 '19 at 16:25

## 1 Answer

Simply put, you cannot expand out the limit as you have done and then recombine, when the parts tend to infinity.

To see what I mean, consider the following reversed example...

\begin{align} 1 = \lim_{x\to0} \frac{x}x &= \lim_{x\to0} \frac{x+1}x - \lim_{x\to0} \frac1x\\ &\neq\lim_{x\to0}\frac{x+1}x(1-x) - \lim_{x\to0} \frac1x\\ &= \lim_{x\to0} \frac{1-x^2}x - \lim_{x\to0} \frac1x\\ &= \lim_{x\to0} \frac{1-x^2 - 1}x\\ &= \lim_{x\to0} \frac{-x^2}x = 0 \end{align}

By looking at the step with the $$\neq$$, we can see what is going on, here... $$\lim_{x\to 0} \frac{x+1}x(1-x) = \lim_{x\to 0} \frac{x+1}x - \lim_{x\to 0}\ (x+1)$$ And so, multiplying $$\frac{x+1}x$$ by $$1-x$$, which has a limit of 1, has the effect of subtracting $$\lim_{x\to 0} (x+1) = 1$$ from the limit, thus changing the result.

The same applies here, albeit in a less obvious form. In particular, notice that $$\frac{\log(1+x)}x \approx 1\color{red}{-\frac{x}2}+O(x^2)$$ If you examine what happens if you add this extra term (in red) to your expression, you end up with the right result.

On a more general note, splitting limits only works if the limits of the parts exist. This applies to both addition and multiplication splitting. For example, $$\lim_{x\to0} \left(\frac1x-\frac1x\right) \neq \lim_{x\to 0} \frac1x - \lim_{x\to 0} \frac1x$$ and $$\lim_{x\to 0} \left(\frac1xx\right) \neq \lim_{x\to 0} \frac1x \times \lim_{x\to 0} x$$

• @peterag - Oops, sorry about that. Fixed. – Glen O Feb 10 '19 at 13:16