How do you find the limit of #[ln(6x+10)-ln(7+3x)]# as x approaches infinity?

Question

Isabella Ackerman · Accepted Answer

Convert to a fraction and then use L'Hôpital's rule. Answer: #ln(2)#

Given: #lim_(xtooo)[ln(6x + 10) - ln(7 + 3x)]#

Use the identity #ln(a) - ln(b) = ln(a/b)#:

#lim_(xtooo)ln((6x + 10)/(7 + 3x)) = #

#ln(lim_(xtooo)(6x + 10)/(7 + 3x))#

Because the above evaluated at the limit is an indeterminate form, #oo/oo#, we should use L'Hôpital's rule :

Compute the derivative of numerator:

#(d(6x + 10))/dx = 6#

Compute the derivative of the denominator:

#(d(7 + 3x))/dx = 3#

Make a new fraction:

#ln(lim_(xtooo)(6/3))#

This can be evaluated at #oo#:

#ln(2)#

Therefore, the original limit goes to the same value:

#lim_(xtooo)[ln(6x + 10) - ln(7 + 3x)] = ln(2)#

Emilia Aguilar · Answer

undefined To find the limit of [ln(6x+10)-ln(7+3x)] as x approaches infinity, we can simplify the expression using logarithmic properties. By applying the quotient rule of logarithms, we can rewrite the expression as ln((6x+10)/(7+3x)).

As x approaches infinity, the terms with the highest degree (6x and 3x) dominate the expression. Therefore, we can ignore the constants (10 and 7) and simplify the expression further to ln(6x/3x).

Simplifying this gives us ln(2), which is a constant. Thus, the limit of [ln(6x+10)-ln(7+3x)] as x approaches infinity is ln(2).