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As you can already see im struggling quite a bit with limits involving the square root! heres a lot of problems!

$$ \lim_{x\rightarrow 1^-} \frac{\sqrt{x^2 -x}}{(x-x^2)} \\ \lim_{x\rightarrow\infty} \frac{x\sqrt{x+1}(1-\sqrt{2x+3})}{(7-6x+4x^2)} \\ \lim_{x\rightarrow-\infty} (\sqrt{x^2 +2x} - \sqrt{x^2-2x}) \\ \lim_{x\rightarrow\infty} (\sqrt{x^2 +2x} - \sqrt{x^2-2x}) \\ \lim_{x\rightarrow\infty} \frac{1}{\sqrt{x^2 -2x} -x} $$ 6) Parking in a certain parking lot costs $\$1.50$ for each hour or part of an hour. Sketch the graph of the function $C(t)$ representing the cost of parking for $t$ hours. At what values of $t$ does $C(t)$ have a limit? Evaluate $\lim_{t\rightarrow t_0^-} C(t)$ and $\lim_{t\rightarrow t_0^+} C(t)$ for an arbitrary number $t_0 > 0$

I would consider myself pretty decent at evaluating limits, but there is some limits problems that has caused some trouble and it would be great if someone could point me in the right direction! tips/advice/solutions greatly appreciated :D

edit: I forgot the denominator on question 2, but should be fixed now, and btw, question: 2,3,4 are marked more difficult in the book. GL

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  • $\begingroup$ You have unbalanced parentheses in limit #2. $\endgroup$ – David H Jun 28 '14 at 14:10
  • $\begingroup$ And unless I'm mistaken, limits #3 and #4 seem to be the same question. $\endgroup$ – David H Jun 28 '14 at 14:12
  • $\begingroup$ @DavidH, 3,4 are different. One is at $\infty$, the other is at $-\infty$ $\endgroup$ – Priyatham Jun 28 '14 at 14:16
  • $\begingroup$ hmm yeah!, I think the answer should be the same tho, except one u get negative answer and the other positive? $\endgroup$ – asdf123 Jun 28 '14 at 14:18
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    $\begingroup$ @asdf123 Would you please use latex? $\endgroup$ – Priyatham Jun 28 '14 at 14:19
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I'll evaluate limit #4 and then #3. If you understand the technique here, you should easily be able to apply it limit #5, which I therefore leave as an exercise to you.

The standard technique for problems like #4 (and #5) is to multiply the function by a fraction equal to $1$ that rationalizes the numerator (for #5 you'll want to rationalize the denominator instead of the numerator):

$$\begin{align} L_4&=\lim_{x\to\infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-2x}\right)\\ &=\lim_{x\to\infty}\left(\frac{\sqrt{x^2+2x}-\sqrt{x^2-2x}}{1}\cdot\frac{\sqrt{x^2+2x}+\sqrt{x^2-2x}}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\right)\\ &=\lim_{x\to\infty}\frac{4x}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\\ &=\lim_{x\to\infty}\frac{4}{\sqrt{1+\frac{2}{x}}+\sqrt{1-\frac{2}{x}}}\\ &=\frac{4}{\sqrt{1}+\sqrt{1}}=2. \end{align}$$

Limit #3 is just like limit #4 except we're evaluating the limit as $x\rightarrow -\infty$ instead of $x\rightarrow\infty$. Substituting $x=-u$, we can find the value of $L_3$ by relating it to $L_4$, which we've already found, instead of doing essentially the same work all over again:

$$\begin{align} L_3&=\lim_{x\to -\infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-2x}\right)\\ &=\lim_{u\to\infty}\left(\sqrt{u^2-2u}-\sqrt{u^2+2u}\right)\\ &=-\lim_{u\to\infty}\left(\sqrt{u^2+2u}-\sqrt{u^2-2u}\right)\\ &=-L_4=-2. \end{align}$$

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