Fdtxjr

I need to find lateral limits of this one $$lim_x to 0xe^frac1x$$ I tried and I got that when $x$ is smaller than $0$ the limit is $0$. But what do I do when $x$ is bigger than $0$?

Is it allowed to use Taylor series in this question? If so, then we have
$$lim_xrightarrow0^+xcdot e^frac1xqquadqquadqquadqquad$$
$$quad= lim_xrightarrow0^+xcdotleft(1+frac1x+frac12!x^2+frac13!x^3+cdotsright)$$
$$= lim_xrightarrow0^+left(x+1+frac12!x+frac13!x^2+cdotsright) ,$$
$$= +inftyqquadqquadqquadqquadqquadquad $$

This is true because every fraction with $x$ in its denominator tends to infinity.

Another approach that only involves differentiation and a little analysis:

$left.right.$

Let $,f(y)=e^y, g(y)=(y^2)/2$, then we have $$f'(y)=e^y, textand ,g'(y)=y$$
Now differentiate again and we get
$$f''(y)=e^y, textand ,g''(y)=1$$
When $,y>0$, $ e^y>1,$ and that means $,f''(y)>g''(y)$.

Also, $ f'(0)=1>0=g'(0)$, so $,f'(y)>g'(y),$ for all $,y>0$

Again, since $ f(0)=1>0=g(0)$, so $,f(y)>g(y),$ for all $,y>0$

So far we have proved that if $,y>0,$ then $,e^y>(y^2)/2$

$left.right.$

Now back to the limit, and we have that

$$lim_xrightarrow0xe^frac1x=lim_yrightarrowinftyfrace^yygeqlim_yrightarrowinftyfrac(y^2)/2y=infty$$

And that is the "exponential dominating a polynomial" in Zahbaz's answer

Are you permitted to rely on an exponential dominating a polynomial in the limit? These limits may look more approachable by transforming $xto frac1y$ such that as $xto0^pm$, $ytopminfty$.

$$lim_x to 0^+xe^frac1x=lim_y to inftyfrace^yy = infty$$

$$lim_x to 0^-xe^frac1x=lim_y to -inftyfrace^yy = 0$$

The last step would be justified with a Taylor expansion, but I'm unclear if that is in your toolkit. You could use an argument like this for the first limit,

$$frace^yy > frace^ye^y/2$$

so

$$lim_y to infty frace^yy > lim_y to inftyfrace^ye^y/2 = lim_y to inftye^y/2=infty$$

$lim_xto 0+𝑥𝑒^1/𝑥= 0 cdot infty$ which is an indefinite form.

We could apply L'Hopital Rule but it can only be done if we have a $0/0$ or $infty/infty$ indefinite forms.
So rewrite $lim_𝑥to0+𝑥𝑒^1/𝑥$ as
$$
lim_𝑥to 0+frac𝑒^1/𝑥frac1x
$$
by putting $x$ in the denominator as $1/x$. Then plug in the limit as $𝑥to 0+$, we get the $infty/infty$ form. Now use L'Hopital's Rule.

1. Take the derivatives of numerator and denominator. We get
   $$
   lim_𝑥to 0+frac-frace^1/xx^2-frac1x^2text the negatives and $x^2$ cancel
   $$
   


2. Simplify, then plug in the limit as $𝑥to0+$. We get
   $lim_𝑥to0+ e^1/x=infty$!!!!

A tip: graph the numerator and denominator separately so you can see the trend