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$$\lim_{x\to\infty}\bigg({\frac{e^x-x-1}{e^x-\frac{1}{2}x^2-x-1}}\bigg)$$
I don't know how to solve this limit, but I have an idea and I want to see if it is right to proceed like this. I am thinking that $e^x$ is increasing faster than $x^2$ and $x$ so maybe this limit is equal to $1$ for this reason? Can I do that? "Ignore" the rest of terms just like I would do in case of limits of fractions when both the numerator and denominator are polynomials?

Radu Gabriel
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4 Answers4

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You are right...but it is usually expected to add an explanation, for example:

$$1=\frac{e^x-x-1}{e^x-x-1}\le\frac{e^x-x-1}{e^x-\frac12x^2-x-1}\le\frac{e^x}{e^x-\frac12x^2-x-1}=\frac1{1-\frac{x^2}{2e^x}-\frac x{e^x}-\frac1{e^x}}\xrightarrow[x\to\infty]{}1$$

so now just apply the squeeze theorem...

DonAntonio
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  • I tried factoring and doing some other things that did not work just so that I would have a better explanation than just "intuition". I know that intuition didn't help me much with math, ESPECIALLY with limits. For example intuition tells me that $1^\infty$ is $1$ and $\frac{0}{\infty}$ is $0$, but I learned that it's not true and I would rather ask and get an explanation from people that are acquainted to the subject instead. – Radu Gabriel Jan 27 '20 at 11:04
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$$L=\lim_{x\rightarrow \infty}\frac{1-(1+x)e^{-x}}{1-(1+x+x^2/2)e^{-x}} =1.$$ Note that $\lim_{x\rightarrow \infty} x^n e^{-x}=0, n \in I^+$

Z Ahmed
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Use asymptotic equivalents:

Since any polynomial $p(x)$ is $o(\mathrm e^x)$ near $+\infty$, we have $\mathrm e^x+p(x)\sim_{+\infty}\mathrm e^x$, so $$ \frac{\mathrm e^x-x-1}{\mathrm e^x-\frac{1}{2}x^2-x-1}\sim_{+\infty} \frac{\mathrm e^x}{\mathrm e^x}=1.$$

Bernard
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We have $\frac{x^n}{e^x} \to 0$ as $x \to \infty$ for all $n \in \mathbb N_0.$

Hence $\lim_{x\to\infty}{\frac{e^x-x-1}{e^x-\frac{1}{2}x^2-x-1}}=\lim_{x\rightarrow \infty}\frac{1-(1+x)e^{-x}}{1-(1+x+x^2/2)e^{-x}} =1.$

Fred
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