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We were given the function below in maths and were told to find the limit as x approaches 0

$\lim_{x\to0}(1+(2x)^2)^{1/x^2}$

the way i did it was just to make all the x's tend towards zero as follows:

$\lim_{x\to0}(1+(0)^2)^{1/0}$

Which i said that means that the limit is undefined. When we went over the answers the professor gave the answer as

$\lim_{x\to0}=e^4$

I am not quite sure how they got this, so could someone point to where i went wrong in my thinking.

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  • $\begingroup$ You wrote $1/0$. $\endgroup$ – Lord Shark the Unknown Apr 28 '18 at 4:17
  • $\begingroup$ I know that and thats how he showed us in previous examples to do it $\endgroup$ – Tom Heeley Apr 28 '18 at 4:18
  • $\begingroup$ Are you saying that a "professor" also wrote $1/0$? $\endgroup$ – Lord Shark the Unknown Apr 28 '18 at 4:20
  • $\begingroup$ Yes I am and thats how he showed that the value was undefined, do you actually know why it equals $e^4$ $\endgroup$ – Tom Heeley Apr 28 '18 at 4:21
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Let $y = \frac{1}{4x^2}$, then $y\to\infty$. $$\lim_{y\to\infty}(1+\frac{1}{y})^{4y}$$ You substituted x by y in the previus step. Now $$\lim_{y\to\infty}(1+\frac{1}{y})^{4y} = \lim_{y\to \infty}((1+\frac{1}{y})^{y})^4=e^4$$ That is becaouse $e$ is defined as $\lim_{y\to \infty}(1+\frac{1}{y})^{y}$

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  • $\begingroup$ You should use proper LaTeX limit notation. Also, you stated in the beginning that $y\to\infty$ but forgot to mention when that happens. $\endgroup$ – ProtectedSource Apr 28 '18 at 4:30
  • $\begingroup$ Thank you so much $\endgroup$ – Tom Heeley Apr 28 '18 at 4:35
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$$\begin{align} & \lim_{x\to0}\left(1+(2x)^2\right)^{\frac1{x^2}}=\lim_{x\to0}\left(1+4x^2\right)^{\frac1{x^2}} \\ & \lim_{x\to\infty}\left(1+\frac4{x^2}\right)^{x^2}=\lim_{x\to\infty}\left(1+\frac4{x}\right)^{x} \\ \end{align}$$

Consider:

$$\begin{align} & e^x=\lim_{n\to\infty}\left(1+\frac{x}n\right)^n \\ \implies& \lim_{x\to0}\left(1+(2x)^2\right)^{\frac1{x^2}}=e^4 \\ \end{align}$$

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  • $\begingroup$ how did you go from $x\rightarrow0$ to $x\rightarrow\infty$ $\endgroup$ – Tom Heeley Apr 28 '18 at 4:32
  • $\begingroup$ I realized that when $x\to0$, then $x^2\to0$, which led me to: as $x\to\infty$, $\frac1{x^2}\to0$. Same result. $\endgroup$ – ProtectedSource Apr 28 '18 at 4:36
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When one theorem says a limit is undefined, like $\infty \cdot 0$, it doesn't guarantee there is not a limit, it just tells you that you need to look closer because you haven't reached a form the theorem can deal with. Compare (with $x \to 0$ from above) $$\lim_{x\to 0} \frac 1x \cdot x^2=\lim_{x\to 0}x=0\\ \lim_{x\to 0} \frac 1x \cdot x=\lim_{x\to 0}1=1 \\ \lim_{x \to 0} \frac 1{x^2} \cdot x=\lim_{x\to 0}\frac 1x=+ \infty$$ All of these are of the form $\infty \cdot 0$ The fact that we have different values on the right side is a demonstration that the basic form $\infty \cdot 0$ is not well enough defined to have an answer. Often you can cancel the product or difference of leading terms analytically, leaving a finite answer.

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I am giving you a general form for solving above type of questions, $\lim_{y\to \infty}(1+\frac{1}{y})^{y}$=$e$ If you clearly see the function, you will realise that the function is a typical example of $1^{(\infty)}$ type of limit. Now to find the power of $e$,
Subtract $1$ from function and multiply it with the exponent of function to get the power of $e$.

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