# To solve the given limit

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?

• 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. – Paramanand Singh May 1 '17 at 7:51
• 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? – Gary Andrews30 May 1 '17 at 7:55
• 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. – 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.
• @GaryAndrews30 Yes, you can get rid of the absolute sign from the very beginning, by considering $-x$ and $x$. – Olivier Oloa May 1 '17 at 8:07