Find the limit using l hospitals rule

Question

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

lim x -> 1 ((7x/x-1) - 7/ln(x))

Answer 1

lim x→1 7x/(x-1)-7/ln(x) = lim x→1 7(xln(x)-x+1)/((x-1)ln(x)) 0/0 case

apply L'Hopital's rule

lim x→1 7(ln(x)+1-1)/(ln(x)+1-1/x) 0/0 case

apply L'Hopital's rule again

lim x→1 7(1/x)/(1/x+1/x²) = 7/(1/(1+1)) = 7/2 = 3.5

Answer 2

use an online limit calculator

and the limit diverges

from the negative side as x approaches 1, the limit diverges to -infinity

if you graph the function, it has two branches, and it may be unclear

what happens at x=1, the right branch approaches +infinity in quadrant I

the left branch approaches -infinity, in quadrant IV

but it's unclear if that's what happens at x=1 or another nearby value of x

using L'Hopital's Rule seems to give 3.5 as the limit

but there may be an error in the following calculations

no guarantees of error free calculations, but it's the basic method:

7x/(x-1) + 7/lnx

put them over a common denominator to combine into one fraction

7xlnx/lnx(x-1) + 7(x-1)/(x-1)lnx

(7xlnx +7x-7)/(xlnx-lnx)

take derivatives, separately of numerator and denominator

(7x/lnx + 7lnx +7)/(1 +lnx - 1/x)

plug in x=1

7/(1-1) = 7/0, still indeterminate

apply L-Hopital a 2nd time

and possibly more times if necessary, until you can plug in x=1 and not get an indeterminate form

(7/x)/(1/x +1/x^2)

plug in x=1

(7/1)/(1/1 +1/1^2)

(7)/(1+1)

= 7/2

limit = 7/2 = 3.5 as x approaches 1

or use a graphing calculator and look at the graph where x=1

this all assumes you didn't mean, as written, 7x/x -1 + 7/lnx instead of 7x/(x-1) + 7/lnx

if by chance you really did mean (7x/x) -1 +7/lnx, then

that = 7-1 +7/lnx

=6 +7/lnx

= (6lnx +7)/lnx

apply L'Hopital's rule to get

(6/x)/(1/x) = 6 = the limit as x approaches 1

you might try 2 or more online limit calculators and see if they agree.

occasionally, they don't. the computer programs aren't perfect

I tried 3 online limit calculators and they all agreed the limit diverges

and the left side limit approaches negative infinity

one limit calculator included a graph which shows the two branches

where the left branch approaches -infinity as x approaches 1