Show that limit of log(2x-3)/2(x-2)=1 as x approaches to 2?

Question

2721 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-15T13:41:36

See below

I'm sure you mean natural logarithm which we write as ln and not log. logx has a base equal to 10!

#Lim_(xrarr2)(ln(2x-3))/(2(x-2))=ln1/(2*0)=0/0#

This is the type of limit 0/0 and that means I can use L'Hospitals rule. (Derivating nominator and denominator)

#[f(g(x))]^'=f^'(g(x))*g^'(x)#

#(ln(2x-3))^'=(ln(2x-3))^'*(2x-3)^'=1/(2x-3)*2#

#(2(x-2))^'=2*(1-0)=2#

#=Lim_(xrarr2)(1/(2x-3))/(2)*2=Lim_(xrarr2)(1/(2x-3))/(cancel2)*cancel2=Lim_(xrarr2)1/(2x-3)=1/(2*2-3)=1/1=1#