# Evaluating $\lim_{x\to 0}\frac{3x^2}{4x^2}$

of $3x^2/4x^2$

Is it undefined or $6/8$?

If we apply the Hospital rule we get $6/8$, so I think it is $6/8$.

It would be undefined if it were $3x^2/4x^3$.

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Don't need to go to Hospital. –  André Nicolas Dec 1 '12 at 20:15

When evaluating limits, remember that the function value at the point is immaterial. What matters is what happens in a "deleted neighborhood" i.e. a neighborhood without the point. So in this case, the function is 3/4 everywhere except at x=0. Hence the limit as x tends to 0 is 3/4. Not undefined just because the function is undefined there.

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Note that for all $x\neq 0$: $$\frac{3x^2}{4x^2} = \frac{3}{4}.$$ Hence then limit is $$\lim_{x\to 0}\;\frac{3x^2}{4x^2} = \lim_{x\to 0}\;\frac{3}{4} = \frac{3}{4}.$$ This of course is the same as $\displaystyle{\frac{6}{8}}$.

The thing is that considering the limit you may assume that $x$ is not equal to zero.

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Here again, while in this problem you CAN use l'Hopital's rule, there really is NO need to use it in this case, and using it doesn't make this limit any easier solve.

We have that for all $x\neq 0$: $\quad \dfrac{3x^2}{4x^2} = \dfrac{3}{4} \left(=\dfrac68\right).$

$$\text{So }\quad \lim_{x\to 0}\;\frac{3x^2}{4x^2} = \lim_{x\to 0}\;\frac{3}{4} = \frac{3}{4}.$$

When taking the limit as $x\to 0$, all that matters is what is happening as $x$ gets VERY VERY close to $0$, not what is happening AT $x = 0$.

It may very well be that for some functions, $\lim_{x\to 0} f(x) = c$ for some value $c$ and that $f(0) = c$, but all that really matters is what happens very near to $x = 0$.

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Why this poor man is named frequently, Hospial? :D –  Babak S. Mar 28 '13 at 9:38

It is $\frac34$ everywhere except at $x=0$, for both positive and negative $x$, so $\frac34$ is the limit at $x=0$.

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...so $\frac34$ is the limit as $x\to 0$, (limit "at" $x$ "=" $0$ may further confuse). –  amWhy Dec 1 '12 at 20:23