Tukukkk

I've been having extreme difficulties with evaluating this limit, I've simply tried everything I can think of, I applied l'hopital's rule twice and I got 1, but it is not included in our curriculum.

$$lim_{x to 0} frac {e^x-sqrt{1+2x}}{x^2}$$

Hint :
$$e^x - sqrt{1+2x} = e^x - sqrt{x^4left(frac{1}{x^4}+frac{2}{x^3}right)}= e^x - x^2sqrt{frac{1}{x^4}+frac{2}{x^3}}$$

Thus, the given limit becomes :

$$lim_{x to 0} left(frac{e^x}{x^2} - sqrt{frac{1}{x^4}+frac{2}{x^3}}right)$$

By the binomial theorem, for small x$$e^x-sqrt{1+2x}approx 1+x+frac12 x^2-(1+x-frac12 x^2)=x^2.$$Thus the limit is $1$ as desired.

If you know how to find a Taylor series, that's the most straightforward way to go, here. If not, let me outline it for you.

The function $f(x)=e^x-sqrt{1+2x}$ is defined on $left[-frac12,inftyright),$ and (infinitely) differentiable on $left(-frac12,inftyright).$ In particular, then, for any $x$ less than $frac12$ away from $0,$ we have that $$f(x)=sum_{k=0}^{infty}frac{f^{(k)}(0)}{k!}x^k,$$ where $f^{(k)}$ denotes the $k$th derivative of $f.$ At first glance, this seems like it wouldn't be useful at all--after all, finding infinitely-many derivatives seems daunting--but fortunately, we don't have to. Instead we need only go partway, by

Taylor's Theorem: Suppose that $n$ is a positive integer, and let $f$ be a real-valued function on a subset of $Bbb R$ that is $n$ times differentiable on an interval $I=(a-R,a+R)$ for some $ainBbb R$ and $R>0.$ Then there is a real-valued function $h_n$ defined on $I$ such that

1. 
   $lim_{xto a}h_n(x)=0,$ and


2. for all $xin I$ we have $$f(x)=h_n(x)(x-a)^n+sum_{k=0}^nfrac{f^{(k)}(a)}{k!}(x-a)^k.$$
   

Why does this help us? Well first, let's calculate the first few derivatives of $f$ to see what happens. Writing $f(x)=e^x+(1+2x)^{frac12},$ we can use the Power and Chain rules of derivatives to quickly see that

$$f'(x)=e^x-frac12(1+2x)^{-frac12}cdot2=e^x-(1+2x)^{-frac12}$$ and $$f''(x)=e^x+frac12(1+2x)^{-frac32}cdot2=e^x+(1+2x)^{-frac32}.$$ We could keep going fairly easily from here and develop an explicit formula for all the other derivatives of $f,$ but there's no need, as you'll see.

By Taylor's theorem with $n=2,$ $a=0,$ and $R=frac12,$ we have that there is a function $h_2$ defined on $left(-frac12,frac12right)$ such that $lim_{xto 0}h_2(x)=0,$ and such that $$begin{eqnarray}f(x) &=& h_2(x)x^2+sum_{k=0}^2frac{f^{(k)}(a)}{k!}x^k\ &=& h_2(x)x^2+f(0)+f'(0)x+frac{f''(0)}2x^2\ &=& h_2(x)x^2+0+0x+frac22x^2\ &=& x^2+h_2(x)x^2end{eqnarray}$$ whenever $|x|<frac12.$ Thus, for $0<|x|<frac12,$ we then have $$frac{f(x)}{x^2}=1+h_2(x).$$ Since $lim_{xto 0}h_2(x)=0,$ then $$lim_{xto 0}frac{f(x)}{x^2}=1,$$ as desired.