Does a Limit Exist if f(5) Does Not?

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If I have a function let's say lim x->5 f(x) that

f(5) Does not exist, because if I give a value of x which equals the limit , there is no anymore limit?


I hope the question is understandble.

Thanks in advance.
 
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The existence of a limit of a function f(x) at a point x=c has nothing to do with whether the function is actually defined there. That's because a control is built into the definition of a limit that prevents x from taking on the value c as we take the limit. That's what the 0<|x-c|<\delta is there for, to ensure that the distance between x and c is strictly positive.
 
Yeah, I like to think of it as making sure that it is approaching c from both sides.

A good example might be \frac{\sqrt{x+1} -1}{x}. The limit exists at c > 0 and the answer is 0 but the you can't just plug that in.

if you let x = -0.1 f(x) is 5.132 and -0.01 f(x) .5013, and -0.001 f(x) is .5001 (from the left), and x is .001, f(x) is 0.4999, x is 0.1, f(x) is 0.4988 (from the right), and so on.

When you study continuity I think it really helps clear things up in regards to the existence of limits.
 
Thank you very much
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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