in exercise i-iii evalute the limit or explain why it does not exist

i) $\displaystyle \lim_{y\to 1}\frac{4-4\sqrt{y}+3}{y^2-1}$

ii) $\displaystyle lim_{x\to 0}\frac{\sqrt{2+x^2}-\sqrt{2-x^2}}{x^2}$

iii) $\displaystyle lim_{x\to 2}\frac{x^4-16}{x^3-8}$

iv) if $\lim_{x\to 2} \frac{f(x)-5}{x-2} = 3,$ find $\lim_{x\to 2}f(x)$

v) suppose $|f(x)| \le g(x)$ for all $x.$ What can you conclude about $\lim_{x\to a} f(x)$ if $\lim_{x\to a} g(x)=0?$ What if $\lim_{x\to a} g(x)= 3?$

any tips/advice/solutions? :d I guess its kinda much to ask but I would rather ask it as one question instead of making a lot, since I dont have any idea how to solve any of these problems:( and for the evaluation of the limit part, question, i, ii, iii I have big trouble with the algebra, and for iii) im pretty much 100% clueless, basically no idea on iv) and v) other than that I know the basics like limit rules and obviously the squeeze theorem aswell, but im not sure how to apply this to these "more complicated" problems. Will be more than grateful with just help on the limit problems!

  • $\begingroup$ Is it $1$ correctly written? $\endgroup$ – mfl Jun 24 '14 at 22:40
  • $\begingroup$ I mean.. all of them are correctly written and the answer to the first one should be -1/2 $\endgroup$ – yayeey123 Jun 24 '14 at 22:41
  • $\begingroup$ If it well written it doesn't exist. When $y\to 1$ the numerator goes to $3$ and the denominator goes to $0.$ $\endgroup$ – mfl Jun 24 '14 at 22:43
  • $\begingroup$ yupp.. that was exactly what I tought, hence why I ask the question here:P the solution however says its -1/2 so idk, I have never ever encountered any wrong solution in this book.. so I think I need to simplify it, idk how tho $\endgroup$ – yayeey123 Jun 24 '14 at 22:45


For the second limit multiply and divide by $\sqrt{2+x^2}+\sqrt{2-x^2}.$ Next, use the identity $(a+b)(a-b)=a^2-b^2.$ That is:

$$\displaystyle lim_{x\to 0}\frac{\sqrt{2+x^2}-\sqrt{2-x^2}}{x^2}=\displaystyle lim_{x\to 0}\frac{(\sqrt{2+x^2}-\sqrt{2-x^2})(\sqrt{2+x^2}+\sqrt{2-x^2})}{x^2(\sqrt{2+x^2}+\sqrt{2-x^2})}$$ $$=\lim_{x\to 0}{2x^2}{x^2(\sqrt{2+x^2}+\sqrt{2-x^2})}=\lim_{x\to 0}\frac{2}{\sqrt{2+x^2}+\sqrt{2-x^2}}$$ $$=\frac{2}{\sqrt{2+0}+\sqrt{2-0}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}.$$

For the third limit factor $x^4-16=(x-2)(x+2)(x^2+4)$ and $x^3-8=(x-2)(x^2+2x+4)$ and simplify.

For the fourth limit, since it exists and the denominator goes to $0$ the same must happen to the numerator.

In the last case use that if $f(x)\le g(x)$ then $\lim_{x\to a}f(x)\le \lim_{x\to a}g(x).$

  • $\begingroup$ well for the second problem the thing that is confusing me is how to multiply that thing with the numerator using (a+b)(a-b)=a^2 -b^2, do I consider the whole thing root(2+x^2) as a? like what do I get in the numerator, I guess is the question $\endgroup$ – yayeey123 Jun 24 '14 at 22:53
  • $\begingroup$ I have edited the answer to write the complete solution of this case. $\endgroup$ – mfl Jun 24 '14 at 22:58
  • $\begingroup$ ahh yeah, I see, thanks a lot! but how did you factor x^4 -16 and x^3 - 8? in the third question? $\endgroup$ – yayeey123 Jun 24 '14 at 23:02
  • $\begingroup$ Use Ruffini's rule en.wikipedia.org/wiki/Ruffini%27s_rule to divide by $x-2.$ Note that these are the problematic factors to be simplified since they vanish at $x=2.$ $\endgroup$ – mfl Jun 24 '14 at 23:06
  • $\begingroup$ ahh ok so its just polynominal division or whatever its called by x-2 since its one of the roots? $\endgroup$ – yayeey123 Jun 24 '14 at 23:08

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