Can someone tell me how to solve this problem and what the answer is.?

Question

Lim(x-ln(x^2+1))
as x approaches infinity

Answer 1

I will give a try ...
As x goes to infinity, x^2 +1 gets really really big. The "+1" part doesn't matter once you get past a few billion billon. So it's just like x^2.

Now ln(x^2) is just 2 times ln(x). And as x gets bigger and bigger, ln(x) get bigger also -- but at a slower rate. So when x get REALLY big, ln(x) is comparatively small. Even 2 times ln(x) is small compared to x.

For example, if x = 10^24, then ln(x) is only 55.26.

So x - 2 ln(x) looks like just x.

The answer is Lim(x - ln(x^2+1)) = x

Answer 2

lim [x - ln(x^2 + 1)]
x -> infinity

This is in the form [infinity - infinity], so it's indeterminate but we cannot use L'Hospital's rule.

One thing to know is that e and ln are inverse of each other. It follows that x can be expressed as ln(e^x). Let's change x into that.
{Note: I'm not going to rewrite x -> infinity over and over again, to save typing, but on paper, that's the way it should be done until the limit is calculated.}

lim ( ln(e^x) - ln(x^2 + 1) )

Now, we can combine these two logs as per the log property
log[base b](a) - log[base b](c) = log[base b](a/c)

lim ( ln [ (e^x) / (x^2 + 1) ] )

We can actually legally bring the limit INSIDE of the log if we wanted to, to get

ln (lim [(e^x) / (x^2 + 1)])

And now, we use L'Hospital's rule on the limit (since we have the form [infinity/infinity]. The derivative of e^x is e^x, and the derivative of x^2 + 1 is 2x, so we have

ln (lim [ (e^x) / (2x) ] )

Now, we STILL have the form [infinity/infinity]. We have to use L'Hospital's rule once more.

ln (lim [ (e^x) / 2 ] )

At this point, we have the form [infinity/2], and we get infinity. The natural log (ln) of something that approaches infinity is infinity, so the limit does not exist.

Answer 3

when you have two terms like that, and both tend to infinity,
x goes to infinity as x goes to infinity
ln(x^2+1) goes to infinity as x goes to infinity

then you have to look at which term approaches infinity FASTER.

for the ln(x^2+1) term, look inside the () parenthesis. you have x^2+1. of the two, 1 is a power of 0 and x^2 is a power of 2, so going towards infinity, x^2 dominates over 1. so you can simplify this term to:

lim [x->inf] ln(x^2+1) = lim [x->inf] ln(x^2)

you can further simplify this (remember law of logs?) to
ln(x^2) = 2*ln (x)


x grows faster than 2 ln (x). a linear term grows faster than a logarithmic approaching infinity, no matter the constant 2. of the remaining equation, the x term dominates. so simplify once again,

lim [x->inf] (x-2*ln (x)) = lim [x->inf] (x)

therefore, the answer is infinity

Answer 4

The limit does not exist because it approaches infinity as x goes to infinity. Since x^2+1 is equal to inifinity as x goes to infinity, and the ln of infinity is infinity, then you get infinity minus infinity, which approaches infinity, aka it does not exist.

Answer 5

well fist u have to tell me the question.