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If $f(x)$ is a function.

Then

Is limit means the value to which $f(x)$ tends , when $x$ tends to a particular constant.

If it is basic rule , then why the same function has two limits as follows.

$\lim\limits_{x\to4} \dfrac{x^2-16}{x-4}$ is originally indeterminate form but after cancelling the common factors new limit will be 8.

And are there any cases such that limit of a function doesn't exist and multiple limit values for same function ?

Why we have to factorize / multiply with some other to get limit value for a function . Why not straight substitution won't give ?

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  • $\begingroup$ When limit will not exist ? $\endgroup$ – Hindu Nov 10 '14 at 16:37
  • $\begingroup$ note that the first limit is undertermined when you intuitively replace $x$ with $4$. The thing is, limits don't replace, only get very very very close. $\endgroup$ – cjferes Nov 10 '14 at 16:43
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The reason for why you can cancel out the term $x - 4$ is that you aren't calculating the value of the function at point $x = 4$. If you were calculating the value of the function at point $x = 4$ you would be dividing by zero, an undefined operation.

If you see the function as a machine where $x$ is inputted in one side while the output is $f(x)$, then comes the question: "What if $x = 4$? Division by $0$?" The answer is that $4$ is not part of the function's domain, which means that $x = 4$ isn't accepted as input and therefore nothing happens. The limit tells you that for $x$ extremely close to $4$, $f(x)$ is extremely close to $8$.

To given an analogy: imagine the function as a box with two holes, one to put things in, another were things come out. The number $4$ is not part of the function's domain, which in this case is something with some shape that won't go in the box, it won't pass through the hole.

You factorize or multiply by a conjugate because you are replacing an expression with something equivalent. In calculus you take advantage of properties from real numbers and algebra, many properties, but you don't prove them.

The limit of a function doesn't exist when the right and left sides approach different quantities. The usual cases are picewise functions and rational functions where there is a vertical asymptote.

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First question: Yes: take for example $e^{1/x}$ as $x\to0$.

Second question: Because the limit is not the value of the function at the point, but the value the function approaches to. When both are the same, we say that the function is continuous.

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Check this out. You will not regret this. This video answers to all your questions: http://www.youtube.com/watch?v=8A_xnWldRRI

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    $\begingroup$ Can you summarize the relevant parts of the video here? Right now, if the video were to be taken down, your answer would be entirely useless. $\endgroup$ – Adam Lear Nov 11 '14 at 14:06
  • $\begingroup$ First 15 minutes of video explans intuituively concept of limits .and whenever it will be taken down give me downvote then ,but first see the treasure $\endgroup$ – godonichia Nov 11 '14 at 14:10

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