# Questions Based On A Couple Of Weird Limits

When evaluating certain limits,i get an answer with which i'am not fully convinced despite following steps which i claim is correct.Answers given by analytic method and L'Hopital's rule differs! I wish i could have used numerical method using MATLAB or other packages to see if that's giving me another different answer.I would like to know the subtlety behind this computation if any.$$1.\lim_{x\to1}\frac{x^2 - 1}{x + 1}$$ $$=\lim_{x\to1}\frac{(x+1)(x-1)}{x+1} = 0$$

When i use L'Hopital's rule for the above limit i.e $$\frac{\frac{\mathrm{d} }{\mathrm{d} x}(x^2-1)}{\frac{\mathrm{d} }{\mathrm{d} x}(x+1)}= \frac{2x}{1} = 2$$

I get different answers when evaluating $$\lim_{x\to2}\frac{x^3-8}{x-2} = \lim_{x\to2}\frac{(x-2)(x^2+2x+4)}{(x-2)}=4+8+4=16$$ Using L'Hopital's rule$$\frac{\frac{\mathrm{d} }{\mathrm{d} x}(x^3-8)}{\frac{\mathrm{d} }{\mathrm{d} x}(x-2)}=\frac{3x^2}{1}=12$$(after taking the limit)

Edit:I realised that L'Hopital's rule can't be applied for the first one as it's not in indeterminate form!But what about the second one?

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L'Hopital's rule can only be used on indeterminate forms – mixedmath Jan 4 '13 at 6:32
I aggree.What about the second one? – alok Jan 4 '13 at 6:34
You mis-evaluated the numerator. $2^2 + 2\cdot 2 + 4 = 12 \not = 16$. – mixedmath Jan 4 '13 at 6:38

L'Hopital's rule is not applicable for $$\lim_{x \to 1} \dfrac{x^2-1}{x+1}$$ since it is not of the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$.
For the second one, your evaluation is incorrect. $$\lim_{x \to 2} \dfrac{x^3-8}{x-2} = \lim_{x \to 2} (x^2 + 2x + 4) = 2^2 + 2 \times 2 + 4 = 4 + 4 + 4 = 12 \,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ You could apply L'Hospital rule for this one since it is in indeterminate form i.e. $$\dfrac{x^3-8}{x-2}$$ evaluated at $x=2$ is of $\dfrac{0}{0}$ form.
Evaluating by L'Hospital's rule, gives us $$\lim_{x \to 2} \dfrac{x^3-8}{x-2} = \lim_{x \to 2} \dfrac{\dfrac{d(x^3-8)}{dx}}{\dfrac{d(x-2)}{dx}} = \lim_{x \to 2} \dfrac{3x^2}{1} = 12$$ which is consistent with $(\star)$