Question

help

Answer 1

Compute

Answer 2

\[\lim_{x \rightarrow 0} (\frac{ 2x^{2}-3x+4 }{ x } +(\frac{ 5x-4 }{ x }))\]

Answer 3

Note, you can add the two fractions as they have a common denominator correct?

Answer 4

OH! WOW okay now I feel dumb

so it's like this right?

so when I did that this is what I get

\[\frac{ 2x^{2}-3x+4+5x-4 }{ x } = \frac{ 2x^{2}+2x+4 }{ x }\]


\[\lim_{x \rightarrow 0}~\frac{ 2x^{2}+2x }{ x } \]


\[\frac{ 2x(x+1) }{ x } = \lim_{x \rightarrow 0}~ 2(x+1) = 2(0+1) = 2 \]

hey what is the purpose behind limits anyway, it just doesn't seem to make sense to me or at least concept wise.

is it suppose to tell us if a function is continuous?

Answer 5

Also like based on that problem i can cancel out x in the denominator but what if you can't do that. how you appoach doing something more complicated than that?

Answer 6

Well done :)

Yes, in a sense, it's quite simple actually but isn't taught with context very well. So if you have something undefined, what the limits tell us is when we come very close to the function we get a specific number. So your limit tells us when we come very close to 0 you get 2. Note your limit exists because it comes to 2 from both the left and right side.

What you're asking is called is indeterminate form, something such as \[\frac{ 0 }{ 0 }\] for example if you had \[\lim_{x \rightarrow 0} \frac{ sinx }{ x }\] which would give you the case of 0/0 and there is a nice little trick to deal with these limits, maybe you can figure it out :P

Answer 7

I always found this very useful: http://betterexplained.com/articles/an-intuitive-introduction-to-limits/

Here is an intuitive way to think of it, I think you will enjoy this!