Question

What is the limit of...

Answer 1

this isn't multiple choice

Answer 2

?

Answer 3

it is indeterminate form, apply L'Hopital's Rule to evaluate...

Answer 4

$$
\large\\
\lim_{x\to0}\cfrac{\sqrt{ x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}
$$
Use the power law over and over:
$$
=\sqrt{\lim_{x\to0}\cfrac{ x+\sqrt{x+\sqrt{x}}}{x+1}}\\
=\sqrt{\cfrac{ \lim_{x\to0}x+\lim_{x\to0}\sqrt{x+\sqrt{x}}}{\lim_{x\to0}x+1}}\\
=\sqrt{\cfrac{ 0+\sqrt{\lim_{x\to0}x+\lim_{x\to0}\sqrt{x}}}{0+1}}\\
=\sqrt{ 0+\sqrt{0+\sqrt{\lim_{x\to0}x}}}\\
=\sqrt{ 0+\sqrt{0+\sqrt{0}}}\\
=0
$$
Does this make sense?

Answer 5

limit approaches to infinity...

Answer 6

Ahh yes. Thanks. Same idea but must use L'Hopitals when getting \(\cfrac{\infty}{\infty}\):

Answer 7

\[\large \begin{align} \\

\lim_{x\to \infty}\cfrac{\sqrt{ x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}} &= \lim_{1/x \to 0}\cfrac{\sqrt{ x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}} \\~\\

&= \lim_{1/x \to 0}\cfrac{\sqrt{ 1+\sqrt{1/x+\sqrt{1/x^3}}}}{\sqrt{1+1/x}} \\~\\
&= \cfrac{\sqrt{ 1+\sqrt{0+\sqrt{0}}}}{\sqrt{1+0}} \\~\\
&= 1
\end{align}\]

Answer 8

In short :

\[\large \begin{align} \\

\lim_{x\to \infty}\cfrac{\sqrt{ x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}} &= \lim_{x \to \infty }\cfrac{\sqrt{\mathcal{O(x)}}}{\sqrt{\mathcal{O(x)}}} \\~\\

&= 1
\end{align}\]

Answer 9

well it seems valid to me... I'm impressed... i learned on that... is it always applicable to indeterminate forms of \(\frac{\infty}{\infty}\)?

Answer 10

yes if the degrees of numerator and denominator are same
(think of horizontal asymptote of a fractional function : f = g/h)

Answer 11

when the degrees are equal, the function approaches to \(1\) as \(x \to \infty \)

Answer 12

ok got your point... thanks really... i learned on that....

Answer 13

*the function approaches to `ratio of leading terms` as \(x \to \infty \)

Answer 14

np :) example :
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