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Mangoes

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## Homework Statement

[itex]

\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - (1 + x)}{x^n}

[/itex]

where n is a positive integer

## The Attempt at a Solution

The numerator approaches 0 as x approaches 0 from the right. The denominator approaches 0 with whichever positive integer, n. This gives an indeterminate form, so I can apply L'Hopital's Rule:

[itex]

\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - 1}{nx^{n-1}}

[/itex]

Now, here's where I'm not too sure...

The numerator will still approach 0. The denominator will still approach 0 as x approaches 0, it's just being multiplied by some constant. If I apply L'Hopital's Rule again:

[itex]

\displaystyle\lim_{x\rightarrow 0^+} \frac{e^x}{n(n-1)x^{n-2}}

[/itex]

The numerator now reaches 1 as x approaches 0.

I'm unsure of how to evaluate the limit because I'm unsure of how to treat the denominator.

If n was some large positive integer, I'd just get (1/0) as x approaches 0 which would yield to infinity... But if n had been some very small integer, the denominator would have eventually simplified to some constant term and evaluating the limit would not give infinity.

How do I actually go on to evaluate the limit without assigning a value for n?

Would it be legal to just specify the two specific cases (n = 1 and n = 2) where the limit doesn't evaluate towards infinity and then just give the general case for n > 2?

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