Question

$li{m}_{x\to 0}\left(\frac{3x+|x|}{7x-5|x|}\right)$=___

• A. $\frac{1}{6}$
• B. $\frac{3}{7}$
• C. 2
• D. does not exist

Answer 1

The correct option is D

does not exist

${lim}_{x\to 0^{-}}\left(\frac{3x+|x|}{7x-5|x|}\right)={lim}_{h\to 0}\left(\frac{3(0-h)+|0-h|}{7(0-h)-5|0-h|}\right)={lim}_{h\to 0}\left(\frac{-3h+h}{-7h-5h}\right)={lim}_{h\to 0}\left(\frac{-2h}{-12h}\right)=\frac{1}{6}{lim}_{x\to 0^{+}}\left(\frac{3x+|x|}{7x-5|x|}\right)={lim}_{h\to 0}\left(\frac{3(0+h)+|0+h|}{7(0+h)-5|0+h|}\right)={lim}_{h\to 0}\left(\frac{3h+h}{7h-5h}\right)={lim}_{h\to 0}\left(\frac{4h}{2h}\right)=2L.H.L\ne R.H.L.$
Hence, the limit does not exist.