The value of $\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}$ I want to find the value of
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}.$$
Since $x \rightarrow + \infty$, I only consider the value of the function for $x \ge 0$, i.e.
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}.$$
Noticing that the numerator and denominator both approaches $+\infty$ as $x$ approaches $+\infty$, I tried using L'Hôpital's Rule. However, it makes things worse. I did check at Wolfram Alpha, and found out that the value of the limit is $\infty$.
I then noticed that both the numerator and the denominator has powers of some function of $x$. Taking natural logarithm of the numerator and denominator, we have $$\dfrac{\ln e^{x^2}}{\ln 10^{x}}=\dfrac{x^2 \ln e}{x \ln 10}=\dfrac{x}{\ln 10}.$$
This seems really silly, but would I be justified if I say that $$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}} = \lim_{x\to +\infty}\dfrac{x}{\ln 10}= + \infty?$$
If this is incorrect, kindly give me a hint to tackle this problem.
Thank you for your time.
 A: You can take the natural logarithm of the fraction and use your idea.
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}=e^{ \lim_{x\to +\infty} \ln\dfrac{e^{x^2}}{10^{x}}}$$
A much simpler solution is to observe that $10 <e^3$ and hence
$$ \dfrac{e^{x^2}}{10^{x}}> \dfrac{e^{x^2}}{e^{3x}}$$

Added: To clarify why I can interchange the exponential and the limit.
Since $\lim_{x\to +\infty} \ln\dfrac{e^{x^2}}{10^{x}} = \infty$, for each $M>0$ there exists some $N>0$ so that $x>N$ implies $f(x) >\ln (M)$. Then for $x >N$ we have 
$$e^{f(x)} >e^{\ln(M)} =M \,.$$
Thus, for each $M>0$ there exists some $N>0$ so that $x>N$ implies $e^{f(x)} >e^{\ln(M)} =M$. This shows that 
$$\lim_{x\to +\infty}e^{ \ln\dfrac{e^{x^2}}{10^{x}}} = \infty  \,.$$
Had $\lim_{x \to \infty} f(x)$ been finite, to show that 
$$\lim_{x \to \infty} e^{f(x)}=e^{\lim_{x \to \infty} f(x)}$$
you'd need to use the continuity of $e^x$ and the following Theorem, which can be typically found in all Calculus/Analysis textbooks:
Theorem If $\lim_{x \to a}f(x)=b$ and $g$ is continuous at $x=b$, then
$$\lim_{x \to a} g(f(x)) = g(b) \,.$$
This Theorem is used to prove that composition of continuous functions is continuous, and it typically be found just before that.
A: Set $k=ln6$, then $10^x=e^{kx}$. The limit you want is $\lim\limits_{x \to \infty}e^{x^2-kx}$. With this your limits can be easily calculated.  
A: The original problem could be handled more directly, but to pick up after what you have done so far: If you have shown that $\lim_{x\to\infty}\frac{\ln(f(x))}{\ln(g(x))}=\infty$, then you have shown that for any $M$, there is an $x_M$ such that for any $x>x_M$, that $$\ln(f(x))>M\cdot\ln(g(x))$$ This implies $$f(x)>g(x)^M$$ and that $$\frac{f(x)}{g(x)}>g(x)^{M-1}$$ Now if $g(x)$ itself approaches $\infty$, then it suffices to settle for any $M>1$ and you have your desired result.
A: You can note that the function $10^x$ is increasing less rapidly than $e^{x^2}$ so that the limit is absolutely $\infty$
