For each of the following find the limit with clear justifications if it exists, or explain why it does not exist.

$(i)$ $\displaystyle \lim_{x \to 0}\frac{\ln(1+x)}{x}$

$(ii)$ $\displaystyle \lim_{x \to 0}\frac{\ln(1+|x|)}{x}$

$(iii)$ $\displaystyle \lim_{x \to 0}\frac{\ln(1+|x|)}{x}\:\sin x$

Question (i) can be solved by using the L'Hôpital's law, I was able to compute that the given limit equals to one. Generally to prove whether a limit exists, would be to use the definition. Should I first compute the limit of the questions manually and then utilise the definition to prove the same? Also I am not aware as to how to go about solving the limits with the modulus signs, is there a particular way to go about doing the same?

  • $\begingroup$ Only the second limit does not exist because left and right limits are different. Other limits exist and first one is 1 and last one is 0. $\endgroup$ – Paramanand Singh May 1 '17 at 7:51
  • $\begingroup$ Why doesn't the second limit exist? Wouldn't the LHL be ln(1-x) and RHL be ln(1+x) And then using the L'Hôpital's rules would end up giving the same value for both intervals? $\endgroup$ – Gary Andrews30 May 1 '17 at 7:55
  • $\begingroup$ You won't get same values for left and right because of the denominator $x$. The numerator changes to $\log(1+x)$ or $\log(1-x)$ based on right and left of $0$, but denominator remains $x$ only. If you apply L'Hospital's Rule you will get 1 for right and -1 for left. Try and see. $\endgroup$ – Paramanand Singh May 1 '17 at 8:13

Hint. Concerning the second limit, one may write, as $x \to 0$, $$ \frac{\ln(1+|x|)}{x}=\frac{\ln(1+|x|)}{|x|}\cdot \frac{|x|}{x} $$ then consider separately $x \to 0^-$ and $x \to 0^+$ using the first limit.

Concerning the last limit, one may observe that $$ \left|\frac{\ln(1+|x|)}{x}\cdot \sin x\right|\le \frac{\ln(1+|x|)}{|x|}\cdot \left|\sin x\right|\le\frac{\ln(1+|x|)}{|x|} $$ then one may use the first limit.

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  • $\begingroup$ Is there a true purpose for writing it that way? Couldn't the absolute sign directly be evaluated and written as -x in the case of the LHL and +x in the case of the RHL or is that a wrong approach? $\endgroup$ – Gary Andrews30 May 1 '17 at 8:05
  • $\begingroup$ @GaryAndrews30 Yes, you can get rid of the absolute sign from the very beginning, by considering $-x$ and $x$. $\endgroup$ – Olivier Oloa May 1 '17 at 8:07

(1) when expansion of ln(1+x) is divided by x and x tending to zero we will get its limit as 1 (2) left hand limit and right hand limit of the function is different hence limit doesn't exist (3) limit exist and using algebra of limit it will be 1 as sinx makes the function tending to same limit from left as well as right hand side

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