[Stuck]: Final step of solving limit, Calculus I

Question

Calculate the value of $k$ such that the following limit has a finite solution, $L$ such that $L \ne 0$:

$$\lim_{x\rightarrow0} \frac{(e^{x^2}-x^2-1)(\cos(x)-1)}{x^k}$$

I use the Taylor Series expansions of $e^x$ and $\cos(x)$ and simplify the above expression to the following:

$$\lim_{x\rightarrow0} \frac{-\frac{1}{4}x^6+(\frac{1}{48}-\frac{1}{12})x^8+(\frac{1}{144})x^{10}}{x^k}$$

Now I have a mental roadblock. For $k<6$ the above limit goes to zero and for $k>6$ this expression should diverge but in my head it goes to zero again.... I am trying to think of a simple example to convince myself but can't. Can someone please help me understand this?

Thanks.

Answer 1

You are almost there

$$\lim_{x\rightarrow0} \frac{-\frac{1}{4}x^6+(\frac{1}{48}-\frac{1}{12})x^8+(\frac{1}{144})x^{10}}{x^k}=\\ =\lim_{x\rightarrow0} -\frac{1}{4}x^{6-k}+(\frac{1}{48}-\frac{1}{12})x^{8-k}+(\frac{1}{144})x^{10-k} $$

If any of the exponents is negative, then the limit goes to infinity, so you must have $k\le 6$. Also if all exponents are (stricly) positive (this happens iff $k<6$) then the limit goes to zero.

Then you must have $k=6$.

for $k>6$ this expression should diverge but in my head it goes to zero again....

Why? If $k>6$, the first (at least) summand goes to infinity.

The standard recipe for limits with polynomials is: "When $x$ goes to infinity, the highest degree term rules; when $x$ goes to zero, the lowest degree term rules". Here, this suggests to factor the ruling term from the fraction and write it as

$$ x^{6-k} ( a + b x^2 +cx^4)$$ for some non-zero $a,b,c$. Then, we quickly see that as $x\to 0$ the factor in parentheses tends to the constant $a$, and we have the three cases: if $k<6$ the left factor tends to zero, if $k>6$ it tends to infinity, if $k=6$ it tends to one (and hence the limit to $a$).

Answer 2

$\cos x-1\sim -\frac{1}{2}x^2$ and $e^{x^2}-x^2-1\sim\frac{1}{2}x^4$, So $k=6$ and $L=-\frac{1}{4}$.

Answer 3

As an alternative derivation, we have that

$$\lim_{x\rightarrow0} \frac{(e^{x^2}-x^2-1)(\cos x-1)}{x^k} =\lim_{x\rightarrow0} \frac{e^{x^2}-x^2-1}{x^{k-2}}\lim_{x\rightarrow0} \frac{\cos x-1}{x^2}=-\frac12\lim_{x\rightarrow0} \frac{e^{x^2}-x^2-1}{x^{k-2}}$$

then recall that

$$e^{x^2}=1+x^2+\frac12x^4+o(x^4)$$

and then for $k=6$

$$\lim_{x\rightarrow0} \frac{e^{x^2}-x^2-1}{x^{4}}=\lim_{x\rightarrow0} \frac{\frac12x^4+o(x^4)}{x^{4}}=\lim_{x\rightarrow0} \frac{\frac12+o(1)}{1}=\frac12$$

and therefore

$$\lim_{x\rightarrow0} \frac{(e^{x^2}-x^2-1)(\cos x-1)}{x^6}=-\frac14$$