Solve indeterminate limit without L'Hopital's Rule I am trying to find $$\lim_{x\to ∞} \frac{x}{\sqrt{x^2+1}}$$
L'Hopital's Rule does not work, as the result will simply be the reciprocal of the original function. I tried to solve by multiplying the function by $\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}$, but that did not get me anywhere. The only help I have found in other questions was to use Taylor Series, but I have not learned that yet and therefore cannot use it.
Any help or hints would be much appreciated
 A: For any $x>0$ we have
$$\frac x{\sqrt{x^2+2x+1}}\le\frac x{\sqrt{x^2+1}}\le\frac x{\sqrt{x^2}}$$
A: Note that for $x > 0$ we have
$$ \frac{x}{\sqrt{x^2 + 1}} = \sqrt{\frac{x^2}{x^2 + 1}} = \sqrt{1 - \frac{1}{x^2 + 1}} $$
and so
$$ \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1}} = \sqrt{\lim_{x \to \infty} \left( 1 -  \frac{1}{x^2 + 1} \right) } = 1. $$
A: $$=\frac{1}{\sqrt{1+\frac{1}{x^2}}}$$
A: As x→∞,$\frac{1}{x}$→0,Take$\frac{1}{x}$ as y
Then,
$$\lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1}}=\lim_{y \to 0} \frac{y}{\sqrt{\frac{1}{y^2} + 1}}$$
$$=\lim_{y \to 0} \frac{y}{\sqrt{\frac{y^2+1}{y^2} }}$$
$$=\lim_{y \to 0} \frac{y}{y.\sqrt{1+y^2}}$$
$$=\lim_{y \to 0} \frac{1}{\sqrt{1+y^2}}$$
$$=1$$
A: This is a very simple limit involving polynomials. The general rule is to factor out the largest power of the variable that tends to +$\infty$, in your case x. 
So in this example you have
$$\lim_{x\rightarrow \infty} \frac{x}{\sqrt{x^2+1}} = \lim_{x\rightarrow \infty}
\frac{x}{x} \frac{1}{\sqrt{1+\frac{1}{x^2}}} =  \lim_{x\rightarrow \infty}  \frac{1}{\sqrt{1+\frac{1}{x^2}}} =1$$
Note that this method always works with polynomials. It also works in the case when you know the limit of $x^n f(x)$, or example when $f(x)=exp(x)$ as you know that $exp(x)$ dominates $x^n$ as $x\rightarrow \infty$
