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When is it allowed to substitute the value of a standard limit such as $\lim_{x\rightarrow0} (\sin x)/x=1$ and $\lim_{x\rightarrow0} \ln(1+x)/x=1$ while adding different functions?

For example if we substitute $\lim_{x\to 0} \ln(1+x)=x$ here we get, $\lim_{x\rightarrow0} (x\cos x-\ln(1+x))/x^2=\lim_{x\rightarrow0} (x\cos x-x)/x^2=\lim_{x\rightarrow0} (\cos x-1)/x=0$ However the actual limit evaluated using L'hospitals rule is $1/2$.

Thus it is wrong to replace $\ln(1+x)$ by $x$. Is it ever possible to substitute $\ln(1+x)$ by $x$ while adding functions.

Vignesh
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    This is a common mistake and I have discussed it in detail in this answer: https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Dec 13 '23 at 14:04
  • Another issue is that $\lim_{x\to 0}\log(1+x)=0$ and not $x$ as you claim. – Paramanand Singh Dec 13 '23 at 14:05
  • @ParamanandSingh sorry by saying that lim x→0 log(1+x)=x I meant that while solving question we we use the standard limit lim x →0 log(1+x)/x=1 and replace log(1+x) by x . I didn't notice that I expressed this in the wrong way . Thank you sir – Vignesh Dec 13 '23 at 15:29

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$\lim_{x\rightarrow0} (x\cos x-\ln(1+x))$ is indeed $\lim_{x\rightarrow0} x\cos x - \lim_{x\rightarrow0} \ln(1+x) = \lim_{x\rightarrow0} x\cos x - \lim_{x\rightarrow0} x$, but that's only because the limits of $x\cos x$ and $\ln(1+x)$ exist.

So where did your working go wrong? You had something of the form limit A/B. The manipulations you performed were valid to find out the limit of A. BUT you can only evaluate limit A and limit B separately if limit B exists and is not equal to 0. Otherwise there's no guarantee it'll work.

So if you abide by those rules and you are not evaluating limit of A, then your operations aren't justified because there's no rule that says you can simplify the sum in a limit when there's more to the limit than just the sum.

SuhailSherif
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  • Thank you sir, I am new to calculus and didn't know when we are allowed to take the limit of the numerator – Vignesh Dec 13 '23 at 13:20
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    @Vignesh you'll get the hang of it as you try more. Just try not to replace terms inside a limit with terms that it is not equal to. – SuhailSherif Dec 13 '23 at 17:17