# limits ( the indeterminate forms)

#### romeobernard

Thanks a lot.

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#### MarkFL

Is it:

$$\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)$$ ?

1 person

#### romeobernard

Is it:

$$\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)$$ ?
yes thats exactly the problem.

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#### MarkFL

yes thats exactly the problem.
Okay, what I would do is set:

$$\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)=L$$

And take the natural log of both sides:

$$\displaystyle \ln\left(\lim_{x \to\infty}\left( \left(1+x^2e^x \right) ^{\Large\frac{1}{x}}\right)\right) =\ln(L)$$

Now, because log functions are one-to-one, we can bring it inside the limit:

$$\displaystyle \lim_{x\to\infty} \left(\ln\left( \left(1+x^2e^x \right) ^{\Large \frac{1}{x}}\right) \right) =\ln(L)$$

Now, applying a log property, we may write:

$$\displaystyle \lim_{x\to\infty} \left(\frac{\ln \left(1+x^2e^x\right)}{x} \right) =\ln(L)$$

Now, we have the indeterminate form $$\displaystyle \frac{\infty}{\infty}$$ on the LHS and can apply L'Hôpital's Rule...what do you get?

1 person

#### romeobernard

Okay, what I would do is set:

$$\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)=L$$

And take the natural log of both sides:

$$\displaystyle \ln\left(\lim_{x \to\infty}\left( \left(1+x^2e^x \right) ^{\Large\frac{1}{x}}\right)\right) =\ln(L)$$

Now, because log functions are one-to-one, we can bring it inside the limit:

$$\displaystyle \lim_{x\to\infty} \left(\ln\left( \left(1+x^2e^x \right) ^{\Large \frac{1}{x}}\right) \right) =\ln(L)$$

Now, applying a log property, we may write:

$$\displaystyle \lim_{x\to\infty} \left(\frac{\ln \left(1+x^2e^x\right)}{x} \right) =\ln(L)$$

Now, we have the indeterminate form $$\displaystyle \frac{\infty}{\infty}$$ on the LHS and can apply L'Hôpital's Rule...what do you get?
thanks wait i will rewrite this equation and see if i can get this.

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#### MarkFL

To follow up, using L'Hôpital's Rule, we may write:

$$\displaystyle \lim_{x\to\infty} \left(\frac{\dfrac{2xe^x+x^2e^x}{1+x^2e^x}}{1} \right) =\ln(L)$$

Or:

$$\displaystyle \lim_{x\to\infty} \left(\frac{x^2+2x}{e^{-x}+x^2} \right) =\ln(L)$$

Applying L'Hôpital's Rule again, we get:

$$\displaystyle \lim_{x\to\infty} \left(\frac{2x+2}{2x-e^{-x}} \right) =\ln(L)$$

Or:

$$\displaystyle \lim_{x\to\infty} \left(\frac{2+ \dfrac{2} {x}}{2- \dfrac{1}{xe^{x}}} \right) =\ln(L)$$

Hence:

$$\displaystyle 1=\ln(L)\implies L=e$$

#### greg1313

You can avoid L'Hôpital's Rule if you would show that

$$\displaystyle \displaystyle\lim_{x \to \infty}(e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^\sqrt{x}e^x)^{1/x} \ \ \ \$$**

$$\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^{x + \sqrt{x}})^{1/x}$$

$$\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^{1 + \tfrac{1}{\sqrt{x}}})$$

$$\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ e$$

** You would want to make sure that you can show that $$\displaystyle \ \ \displaystyle\lim_{x \to \infty}\bigg(\dfrac{e^\sqrt{x}e^x}{1 + x^2e^x}\bigg) > 1$$.

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