Question Number 106435 by Study last updated on 05/Aug/20
$${li}\underset{{x}\rightarrow\mathrm{2}} {{m}}\frac{{ln}\left(\mathrm{2}{x}+\mathrm{10}\right)}{{x}−\mathrm{2}}=?? \\ $$
Commented by Dwaipayan Shikari last updated on 05/Aug/20
$${limit}\:{doesn}'{t}\:{exist} \\ $$
Commented by bemath last updated on 05/Aug/20
$$\mathrm{ln}\:\left(\mathrm{14}\right)\:\neq\:\mathrm{0}\:.\:\mathrm{so}\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{rule}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\mathrm{work}. \\ $$
Answered by 1549442205PVT last updated on 05/Aug/20
![Set x−2=t we get lim_(x→2) ((ln(2x+10))/(x−2))= lim_(x→2) ((ln[2(x−2)+14])/(x−2))=lim_(t→0) ((ln(2t+14))/t) = _(Hopital ) lim_(t→0) (2/(2t+14))=(2/(14))=(1/7)](Q106450.png)
$$\mathrm{Set}\:\mathrm{x}−\mathrm{2}=\mathrm{t}\:\mathrm{we}\:\mathrm{get}\:{li}\underset{{x}\rightarrow\mathrm{2}} {{m}}\frac{{ln}\left(\mathrm{2}{x}+\mathrm{10}\right)}{{x}−\mathrm{2}}= \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{ln}\left[\mathrm{2}\left(\mathrm{x}−\mathrm{2}\right)+\mathrm{14}\right]}{\mathrm{x}−\mathrm{2}}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{2t}+\mathrm{14}\right)}{\mathrm{t}} \\ $$$$\underset{\mathrm{Hopital}\:} {=\:\:\:\:}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\frac{\mathrm{2}}{\mathrm{2}\boldsymbol{\mathrm{t}}+\mathrm{14}}=\frac{\mathrm{2}}{\mathrm{14}}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$
Commented by Her_Majesty last updated on 05/Aug/20
$${you}\:{cannot}\:{use}\:{l}'{Hopital}\:{in}\:{this}\:{case}\:{because} \\ $$$${ln}\:\left(\mathrm{2}{t}+\mathrm{14}\right)\:{is}\:{defined}\:{for}\:{t}=\mathrm{0} \\ $$
Commented by 1549442205PVT last updated on 05/Aug/20
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}.\mathrm{I}\:\mathrm{mistaked}.\mathrm{That}\:\mathrm{limit} \\ $$$$\mathrm{lead}\:\mathrm{to}\:\infty\: \\ $$