## Limit at Infinity Problems with Square Roots

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Anonymous
3 months ago

limit of square root of x^2+1 – square root of x+1 as x approaches infinity

Matheno
Editor
3 months ago

As with so many of the problems on this page, we first multiply by the conjugate of the given expression divided by itself:
\begin{align*}
\lim_{x \to \infty}\left(\sqrt{x^2 + 1 } – \sqrt{x+1} \right) &= \lim_{x \to \infty}\left(\sqrt{x^2 + 1 } – \sqrt{x+1} \right) \cdot \frac{\sqrt{x^2 + 1 } + \sqrt{x+1}}{\sqrt{x^2 + 1 } + \sqrt{x+1}} \8px] &= \lim_{x \to \infty}\frac{\left(x^2 + 1\right) + \sqrt{x^2 + 1 }\sqrt{x+1} + \left(- \sqrt{x+1} \right)\sqrt{x^2 + 1 } – (x+1)}{\sqrt{x^2 + 1 } + \sqrt{x+1}} \\[8px] &= \lim_{x \to \infty} \frac{x^2 -x}{\sqrt{x^2 + 1 } + \sqrt{x+1}}\\[8px] \end{align*} Next, we employ the “trick” we discussed on the page “Limits at Infinity,” and divide both the numerator and denominator by the largest power of x that is in the denominator. In this case, that power is \sqrt{x^2} = x. Starting with the last line above: \begin{align*} \lim_{x \to \infty}\left(\sqrt{x^2 + 1 } – \sqrt{x+1} \right) &= \lim_{x \to \infty} \frac{x^2 -x}{\sqrt{x^2 + 1 } + \sqrt{x+1}}\\[8px] &= \lim_{x \to \infty} \frac{\dfrac{1}{x}\left(x^2 -x\right)}{\dfrac{1}{x}\left( \sqrt{x^2 + 1 } + \sqrt{x+1}\right)} \\[8px] &= \lim_{x \to \infty} \frac{x-1}{\sqrt{\dfrac{x^2 + 1}{x^2} } + \sqrt{\dfrac{x+1}{x^2}}} \\[8px] &= \lim_{x \to \infty} \frac{x-1}{\sqrt{1 + \dfrac{1}{x^2}} + \sqrt{\dfrac{1}{x} + \dfrac{1}{x^2}} } \end{align*} Now, \displaystyle{\lim_{x \to \infty} \dfrac{1}{x^2} = 0 } and \displaystyle{\lim_{x \to \infty} \dfrac{1}{x} = 0 }, so \begin{align*} \lim_{x \to \infty}\left(\sqrt{x^2 + 1 } – \sqrt{x+1} \right) &= \lim_{x \to \infty} \frac{x-1}{\sqrt{1 + \cancelto{0}{ \dfrac{1}{x^2}}} + \sqrt{\cancelto{0}{ \dfrac{1}{x}} + \cancelto{0}{ \dfrac{1}{x^2}}} } \\[8px] &= \lim_{x \to \infty} (x-1) = \infty \quad \cmark \end{align*} We hope that helps! Last edited 3 months ago by Matheno Jayshree 6 months ago Please help Matheno Editor Reply to Jayshree 6 months ago We’ve updated the page to provide the solution to this question as Practice Problem #8 above. It’s a good problem, and we hope we’ve been able to help you (and future students) solve it successfully for yourself. Thanks for asking! Jayshree 6 months ago √x+√x+√x/√x please help Last edited 6 months ago by Jayshree Matheno Editor Reply to Jayshree 6 months ago We’ve updated the page to provide the solution to this question as Practice Problem #8 above. It’s a good problem, and we hope we’ve been able to help you (and future students) solve it successfully for yourself. Thanks for asking! Anonymous 1 year ago root of x ( root of x+c – root x ) at lim x tends to infinite Matheno Editor Reply to Anonymous 1 year ago Thanks for asking! (Note that you may need to refresh your browser page for the math below to display properly.) It turns out there are two strategic moves you have to make to find this limit. First, as discussed above, any time we see a two square roots subtracted from each other, we automatically multiply by the conjugate \[1= \frac{\sqrt{x+c} + \sqrt{x}}{\sqrt{x+c} + \sqrt{x}}
So
\begin{align*}
\lim_{x \to \infty}\sqrt{x} \left( \sqrt{x+c} – \sqrt{x}\right) &= \lim_{x \to \infty} \sqrt{x}\left[\left( \sqrt{x+c} – \sqrt{x}\right) \cdot \left( \frac{\sqrt{x+c} + \sqrt{x}}{\sqrt{x+c} + \sqrt{x}}\right) \right] \\[8px]
&= \lim_{x \to \infty}\sqrt{x} \left[ \frac{(x+c) +\sqrt{x+c}\sqrt{x} – \sqrt{x}\sqrt{x+c}-x }{\sqrt{x+c} + \sqrt{x}}\right] \\[8px]
&= \lim_{x \to \infty}\sqrt{x} \left[ \frac{c}{\sqrt{x+c} + \sqrt{x}}\right] \\[8px]
\end{align*}
The other strategic move we need to make is a “trick” we developed earlier : find the largest power of $x$ in the denominator, and then factor it out. Here, that power is $\sqrt{x}$:
\begin{align*}
\phantom{ \lim_{x \to \infty}\sqrt{x} \left( \sqrt{x+c} – \sqrt{x}\right)} &= \lim_{x \to \infty}\sqrt{x} \left[ \frac{c}{\sqrt{x}\left( \sqrt{1+\frac{c}{\sqrt{x}}} + 1\right)}\right] \\[8px]
&= \lim_{x \to \infty}\frac{\sqrt{x}}{\sqrt{x}} \left[ \frac{c}{\sqrt{1+\frac{c}{\sqrt{x}}} + 1}\right] \\[8px]
&= \lim_{x \to \infty} \frac{c}{ \sqrt{1+\frac{c}{\sqrt{x}}} + 1}
\end{align*}
Now, since $\lim_{x \to \infty} \dfrac{c}{\sqrt{x}} = 0,$ when we take the limit we have
\begin{align*}
\phantom{ \lim_{x \to \infty}\sqrt{x} \left( \sqrt{x+c} – \sqrt{x}\right)} &= \frac{c}{ \sqrt{1+\cancelto{0}{ \frac{c}{\sqrt{x}}}} + 1} \\[8px]
\end{align*}
Whew! : )

We hope that helps.

Anonymous
1 year ago

√x√x√x-√x Lim app to infinity is

Matheno
Editor