Evaluate the limit using L'Hospital's rule

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

So when you plug infinity into the original equation, you end up with infinity - infinity. However, 0 is not the answer. So how do I go about solving this? Is L'Hospital's rule necessary? Or is there a rule for when a limit is (infinity - infinity)?

infinity - infinity is not 0, it is undefined. Therefore, L'Hopital's rule is the correct way to approach this.

First, pull out the 2.

2 * lim x -> infinity x(e^(1/x) - 1)

This is still indeterminate infinity * 0.

Let's rewrite this as a fraction:

2 * lim x -> infinity (e^(1/x) - 1)/(1/x)

Lets do a little bit of substitution to make this less ugly. let u = 1/x. As x -> infinity, u -> 0, so we can completely rewrite this in terms of u.

2 * lim u -> 0 (e^u - 1)/u

now lets use L'Hopital's rule and derive the numerator and denominator.

2 * lim u -> 0 (e^u)/1
2 * lim u -> e^u

e^0 = 1

2 * 1 = 2.

No, there is no rule with infinity - infinity. If you want to use l'Hopital's rule, you need to put it in the usual 0/0 or infinity/infinity forms. Here's one way to do it:

2xe^(1/x) - 2x
= 2x(e^(1/x) - 1)
= 2(e^(1/x) - 1) / (1/x)

The numerator tends to 0, as does the denominator, so we may use l'Hoptial's rule. The limit as x approaches infinity will be the same as the limit of:

2(e^(1/x) * -1/x^2) / (-1/x^2)
= 2e^(1/x)
----> 2

So, the limit is 2.

∞ - ∞ is indeterminate form. You need to rewrite equation so that limit becomes indeterminate form 0/0 or ±∞/∞, then you can use L'Hospital's rule

lim[x→∞] (2x e^(1/x) - 2x)
= lim[x→∞] 2x (e^(1/x) - 1)
= lim[x→∞] 2 (e^(1/x) - 1) / (1/x) = 2(e^0 - 1) / 0 = 0/0

Now we can use L'Hospital's rule
= lim[x→∞] 2 ((-1/x²) e^(1/x)) / (-1/x²)
= lim[x→∞] 2 e^(1/x)
= 2 e^0
= 2

Ματπmφm

Much easier without L'Hopital
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Note : lim (x→0) [ (a^x) - 1 ] / x = ln a ...... (1)
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L = lim (x→∞) [ 2x. e^(1/x) - 2x ]

... = 2 • lim (x→∞) [ (e^(1/x)) - 1 ] / (1/x)

... = 2 • lim (h→0) [ (e^h) - 1 ] / h, .............. h = (1/x) → 0

... = 2 • ( ln e ) .................................. from (1)

... = 2 • ( 1 )

... = 2 .................................. Ans.
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