Lim x->infinity ((x+10)/(x-6))^5x

Question

L'hopital is allowed

Answer 1

Given & Find:

L = lim_x→∞( (( x + 10 ) / ( x - 6 ))^5x )

Solution:

= lim_x→∞( e^{5x * ln( ( x + 10 ) / ( x - 6 )} )

= exp( 5 * lim_x→∞( x * ln( ( x+10 ) / ( x-6 )) ) )

To be able to better see how L'Hopital's Rule fits into this problem, I suggest doing this substitution:

x = 1 / n

Turning the limit into:

= exp( 5 * lim_n→0( ln[ ( 1 / n + 10 ) / ( 1 / n + 6 ) ] / n ) )

And then I suggest multiplying the "top and bottom" of the quotient in the natural log 'n', making it:

= exp( 5 * lim_n→0( ln[ ( 1 + 10n ) / ( 1 + 6n ) ] / n ) )

= exp( 5 * lim_n→0( ln[ ( 1 + 10(0) ) / ( 1 + 6(0) ) ] / (0) ) ) = exp( 5 * ln( 1 ) / 0 ) = exp( 5 * 0 / 0 )

Now we can definitively use L'Hopital's Rule w.r.t. (with respect to) n

^d/_dn( ln[ ( 1 + 10n ) / ( 1 + 6n ) ] ) = 1 / [ ( 1 + 10n ) / ( 1 + 6n ) ] * ^d/_dn[ ( 1 + 10n ) / ( 1 + 6n ) ]

= [ ( 1 + 6n ) / ( 1 + 10n ) ] *[ 10 ( 1 + 6n ) - ( -6 ) ( 1 + 10n ) ] / [ ( 1 + 6n )² ]

^d/_dn( ln[ ( 1 + 10n ) / ( 1 + 6n ) ] ) = [ 10 + 60n + 6 + 60n ] / [ ( 1 + 10n )( 1 + 6n ) ]

^d/_dn( ln[ ( 1 + 10n ) / ( 1 + 6n ) ] ) = [ 16 + 120n ] / [ ( 1 + 10n )( 1 + 6n ) ]

L = exp( 5 * lim_n→0( [ 16 + 120n ] / [ ( 1 + 10n )( 1 + 6n ) ] / 1 ) )

= exp( 5 * [ 16 + 120(0) ] / [ ( 1 + 10(0) )( 1 + 6(0) ) ]

= exp( 5 * 16 )

∴ L = e⁸⁰

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!

Answer 2

The is graph shows that this function is unbounded as x -> ∞.

To make sense of this remember that log lim (fx) = lim log (f(x) and look at the log of the given function and l'Hospital's rule will show you another way to see it.