Calculate this limit without l'hospital's rule

Lim Ln(x) / Ln((x^2)-1)
x->+infinity

@John D
Even though limit is indeed 1/2, your 3rd step is invalid, since
ln(x²-1) is NOT ln(x²) - ln(1) -----> ln(x²) - ln(1) = ln(x²/1) = ln(x²) ≠ ln(x²-1)

Recall that ln(a*b) = ln(a) + ln(b)
What you did was ln(a+b) = ln(a) + ln(b), which is incorrect

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First, we'll use a change of variable:
lim[x→∞] ln(x)/ln(x²-1)

Let u = x²-1, x = √(u+1). Then as x→∞, u→∞
Limit becomes
lim[u→∞] ln(√(u+1))/ln(u) = lim[x→∞] 1/2 ln(u+1)/ln(u)
which is the same as lim[x→∞] 1/2 ln(x+1)/ln(x)

Therefore: lim[x→∞] ln(x)/ln(x²-1) = lim[x→∞] 1/2 ln(x+1)/ln(x)

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Now for large positive x,
x² > x²-1
ln(x²) > ln(x²-1) . . . . . . since ln is an increasing function
ln(x)/ln(x²) < ln(x)/ln(x²-1)

lim[x→∞] ln(x)/ln(x²)
= lim[x→∞] ln(x)/(2ln(x))
= lim[x→∞] 1/2
= 1/2

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For large positive x,
2x > x+1
ln(2x) > ln(x+1) . . . . . . since ln is an increasing function
ln(2x)/ln(x) > ln(x+1)/ln(x)
1/2 ln(2x)/ln(x) > 1/2 ln(x+1)/ln(x)
1/2 ln(x+1)/ln(x) < 1/2 ln(2x)/ln(x)

lim[x→∞] 1/2 ln(2x)/ln(x)
= lim[x→∞] 1/2 (ln(2) + ln(x))/ln(x)
= lim[x→∞] 1/2 (ln(2)/ln(x) + ln(x)/ln(x))
= lim[x→∞] 1/2 (ln(2)/ln(x) + 1)
= 1/2 (0 + 1)
= 1/2

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So for large positive x, we have:
1/2 = lim[x→∞] ln(x)/ln(x²) < lim[x→∞] ln(x)/ln(x²-1) = lim[x→∞] 1/2 ln(x+1)/ln(x)
< 1/2 ln(2x)/ln(x) = 1/2

Therefore, by Squeeze Theorem:
lim[x→∞] ln(x)/ln(x²-1) = 1/2

Mαthmφm

ln(x)=yln(x^2-1)
ln(x)/y=ln(x^2-1)
ln(x)/y=ln(x^2)-ln(1)
ln(x)/y=2ln(x)-ln(1)
1/y=(2ln(x)-ln(1))/ln(x)
1/y=2ln(x)/ln(x)-ln(1)/ln(x)
1/y=2-ln(1)/ln(x)

remember that the limit answer is y

so if x goes to plus ∞ ln(1)/ln(x) goes to 0 (you should know that)