I am not able to verify this limit, could someone show a step by step solution to this question?
$$ \lim_{h\to 0} \frac{((h+x)+2)^{3/5}-(x+2)^{3/5}}{h} $$ evaluated at $x=-2$.
I wish to see it derived without using derivatives.
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Don't you want $x=2$? – Ragnar Jul 16 '14 at 19:45
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Don't you have any conditions in h? I guess h should be approaching 0 (you're taking the derivative of $(x-2)^{3/5}$) – S - Jul 16 '14 at 19:46
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2Are you trying to look at $((x-2)^{3/5})'$ evaluated at $x=-2$? – DanZimm Jul 16 '14 at 19:47
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Im sorry was a typo, it was negative where it was suppose to be positive. Could you guys have a look at it now. – ALEXANDER Jul 16 '14 at 19:51
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I have corrected it! – ALEXANDER Jul 16 '14 at 19:54
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If you don't have any conditions in h you just have (if h is not approaching 0): $\frac{h^{3/5}}{h}=h^{-2/5}$ – S - Jul 16 '14 at 19:55
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Off, of course. Limit as h-> approaches zero. I also want to see how its done without using L.H rule by derivatives. – ALEXANDER Jul 16 '14 at 20:00
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@ALEXANDER it is undefined not 0. Please look at my answer – Varun Iyer Jul 16 '14 at 20:01
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The functions is $(x+2)^\frac{3}{5}$, not $(x-2)^\frac{3}{5}$ – Varun Iyer Jul 16 '14 at 20:04
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In the answer manual it states that the limit is -2. In other words I can not verify that it is 0. – ALEXANDER Jul 16 '14 at 20:06
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@ALEXANDER then your problem is written incorrectly. Please update. My solution below is correct. – Varun Iyer Jul 16 '14 at 20:08
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Okey, looked back at it several times now, it is correctly stated. First error Ive found in the text book. – ALEXANDER Jul 16 '14 at 20:10
3 Answers
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Without using derivatives you wish to find $$ \lim_{h \to 0} \frac{((x+h)+2)^{3/5}-(x+2)^{3/5}}{h} \tag{1} $$ at $x = -2$ then we can evaluate $$ \lim_{h \to 0} \frac{((-2+h)+2)^{3/5}-(-2+2)^{3/5}}{h} = \lim_{h \to 0} h^{-2/5} \to \infty $$ Note however that if you wish to find the limit as $x \to -2$ of $(1)$ then you want to consider $$ \lim_{x \to -2} \lim_{h \to 0} \frac{((x+h)+2)^{3/5}-(x+2)^{3/5}}{h} = \lim_{x \to -2} \frac{3}{5}(x+2)^{-2/5} \to \infty $$ but you come to the same issue of getting arbitrarily large values as you get closer and closer to your limit point.
DanZimm
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If you want to evaluate the limit at a particular value of $x$, simply plug in that number and then take the limit. That is, you get
$$\lim_{h\to0}{((h-2)+2)^{3/5}-(-2+2)\over h}=\lim_{h\to0}{h^{3/5}-0\over h}=\lim_{h\to0}{1\over h^{2/5}}$$
which, being of the form $1/0$, is undefined.
Note, I italicized evaluate to emphasize a conceptual difference between $f'(a)$ and $\lim_{x\to a}f'(x)$.
Barry Cipra
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I see that DanZimm posted a virtually identical answer while I was taking my time writing up my own. Great minds and all that.... – Barry Cipra Jul 16 '14 at 20:19
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What I actually did was use telepathy and just have insane typing skills ;P – DanZimm Jul 16 '14 at 20:22
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1@DanZimm, LOL. I wondered who was tapping my brainwaves. Please return any unwanted thoughts. – Barry Cipra Jul 16 '14 at 20:28
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This limit is equal to finding the derivative for a function:
$$f(x) = (x+2)^\frac{3}{5}$$
At $x = -2$
So, $$f'(x) = \frac{3}{5}*(x+2)^\frac{-2}{5}$$
$$f'(-2) = \frac{3}{5*0} = \text{undefined}$$
You are incorrect.
Varun Iyer
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