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# What happens if you try to use l'Hospital's Rule to find the limit? Evaluate the limit using another method.$lim_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}$

## $\lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^{2}+1}} \stackrel{\text { ? }}{=} \lim _{x \rightarrow \infty} \frac{1}{\frac{1}{2}\left(x^{2}+1\right)^{-1 / 2}(2 x)}=\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+1}}{x},$ Repeated applications of 1'Hospital's Rule result in theoriginal limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominatorby $x: \lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^{2}+1}}=\lim _{x \rightarrow \infty} \frac{x / x}{\sqrt{x^{2} / x^{2}+1 / x^{2}}}=\lim _{x \rightarrow \infty} \frac{1}{\sqrt{1+1 / x^{2}}}=\frac{1}{1}=1$

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okay, If we try to use the hospital's rule to evaluate this limit, we see that we end up with something strange, which is that we end up with the same thing that we started with flipped, as you can see. Therefore, we know we have to rewrite the limit and manipulate it in a different way. So usually, when the hospital's rule doesn't work, you have to manipulate the limit. So let's put the whole thing under the square root so the derivative of the top is too accident expertise to exit and then derivative of X squared plus one is two acts. The post one just goes and derivative of zero, which we don't even have to acknowledge. This gives us the squared of one, which is simply one right, because the trucks on top and chips and one cancels

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