# 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

For any $x>0$ we have

$$\frac x{\sqrt{x^2+2x+1}}\le\frac x{\sqrt{x^2+1}}\le\frac x{\sqrt{x^2}}$$

• So your solution works because x is approaching infinity, so the addition of 1 (a constant) does not affect the solution? Or am I misunderstanding? Aug 31, 2016 at 2:07
• @Andrew. For $x>0$ the far LHS is equal to $\frac {x}{x+1}$ and the far RHS is equal to $1$. We can get a sharper estimate for $\sqrt (x^2+1)$ For $x>0$we have $x(1+1/2(x^2+1)<\sqrt (x^2+1)<x(1+1/2x^2).$ Aug 31, 2016 at 8:19
• @Andrew. I ran out of editing time. That should be : For $x>0$ we have $x(1+\frac {1}{2+2x^2})<\sqrt {x^2+1}<x(1+\frac {1}{2x^2}).$ Aug 31, 2016 at 8:27

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

$$\lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1}} = \sqrt{\lim_{x \to \infty} \left( 1 - \frac{1}{x^2 + 1} \right) } = 1.$$

• Nice solution.$\;$ May 12, 2018 at 2:30

$$=\frac{1}{\sqrt{1+\frac{1}{x^2}}}$$

As x→∞,$\frac{1}{x}$→0,Take$\frac{1}{x}$ as y Then, $$\lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1}}=\lim_{y \to 0} \frac{y}{\sqrt{\frac{1}{y^2} + 1}}$$ $$=\lim_{y \to 0} \frac{y}{\sqrt{\frac{y^2+1}{y^2} }}$$ $$=\lim_{y \to 0} \frac{y}{y.\sqrt{1+y^2}}$$ $$=\lim_{y \to 0} \frac{1}{\sqrt{1+y^2}}$$ $$=1$$

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 $x\rightarrow \infty$