Find the limit as x approaches infinite of (3xe^(1/x)-5x^2). Show Work?

Answer 1

lim(3xe^(1/x) - 5x^2) as x →∞ →
lim(3x - 5x^2) as x →∞ = -∞
(the x^2 term dominates)

Answer 2

So you want to solve

lim (3xe^(1/x) - 5x^2)
x -> infinity

Here, your have the form (infinity(1) - infinity, which is definitely an indeterminate form. What you need to do is to make this form into a fraction which L'Hospital's rule can be applied.

Let's factor out x^2. {Note: from now on, I'm *not* going to rewrite "x -> infinity" but on paper it's something you should do}.
Then, we have

lim ( x^2 ([3e^(1/x)]/x - 5]) )

Now, let's give the bracketed terms a common denominator.

lim (x^2 ([3e^(1/x)]/x - 5x/x) )

Merging the two fractions, we have

lim (x^2 ([3e^(1/x) - 5x]/x )

This is now of the form [infinity/infinity], and we use L'Hospital's rule. We use the product rule and chain rule on the numerator. Now, we take the derivative.

lim ( 2x[3e^(1/x) - 5x] + x^2[3e^(1/x) {-1/x^2} - 5] )

Now, we simplify this. Note that the x^2 and -1/x^2 cancel out.

lim ( 6xe^(1/x) - 10x^2 - 3e^(1/x) - 5x^2)

Grouping like terms, we have

lim (6xe^(1/x) - 15x^2 - 3e^(1/x))

lim (6xe^(1/x) - 3e^(1/x) - 15x^2)

Factoring the first two terms,

lim (3e^(1/x) [2x - 1] - 15x^2)

Let's put this under a denominator of x^2 again. This makes the second term -15x^4.

lim ( {3e^(1/x) [2x - 1] - 15}/(x^2) )

Pulling the constant 3 out of the limit,

3 * lim ( {e^(1/x) [2x - 1] - 5} / (x^2) )

Let's use L'Hospital's rule again

3 * lim ( { e^(1/x) (-1/x^2) [2x - 1] + e^(1/x) (2) } / {2x}

Multiplying top and bottom by -x^2,

3 * lim ( {e^(1/x) (2x - 1) - 2e^(1/x)[x^2] } / (-2x^3)

This is as far as I'm going. I might have taken a wrong turn.

Answer 3

Rewrite the equation. Write it as 3x/(e^(-1/x)e^(5x^2)). Then take the limit and use L'Hopital's rule. This will give you the limit as zero, which you can probably tell from your calculator.

Answer 4

graph it and if there is a maximum then calculate it...
I dont believe you can find a limit on infinite... but then again, I am not in your math class...

Answer 5

do your own homework...