Lim x→0 (x + (1/x)) sin(x) Undefined

  • Thread starter Dr-NiKoN
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In summary, as x approaches 0, the expression (x + (1/x))sin(x) approaches 1. This can be shown by breaking the expression into two parts and taking the limit of each part separately. The limit of the first part is 0 and the limit of the second part is 1, resulting in a limit of 1 for the entire expression. This can also be proven using l'Hopital's rule.
  • #1
Dr-NiKoN
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lim_x->0 (x + (1/x) ) sin(x)

As x goes towards 0, wouldn't 1/x go towards inf?
Thus
(x + (1/x)) should go towards inf as x goes towards 0?

sin(x) will go towards 0 as x goes towards 0, since sin(0) is 0.

Wouldn't that leave inf * 0 = undefined (or 0)?
 
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  • #2
Note that your expression is the sum of two other expressions:
[tex](x+\frac{1}{x})\sin(x)=x\sin(x)+\frac{\sin(x)}{x}[/tex]
1. If it can be shown that each of these expressions (on the right-hand side) has a limit as x->0, what's then true about the limit of their sum (i.e, your original expression)?
 
  • #3
lim_x->0 f(x) = x*sinx = 0
lim_x->0 g(x) = sinx/x = 1

lim_x->0 f(x) + g(x) = 0 + 1

lim_x->0 (x + 1/x))sinx = 1

Hm, that was a lot easier than I thought, thanks a lot :)

Edit: How do you do latex?
test:
[itex]\pi[/itex]

nevermind :)
 
  • #4
ah, just l'hopital the second part :-)
 

FAQ: Lim x→0 (x + (1/x)) sin(x) Undefined

What does "Lim x→0 (x + (1/x)) sin(x) Undefined" mean?

The notation "Lim x→0 (x + (1/x)) sin(x) Undefined" represents the limit of a function as x approaches 0. In this case, the function is (x + (1/x)) sin(x). The undefined part means that the limit cannot be determined using traditional methods.

Why is the limit of the function undefined?

The function (x + (1/x)) sin(x) has a singularity at x = 0, meaning that it is undefined at this point. This occurs because the denominator of the fraction (1/x) becomes 0 when x approaches 0, resulting in an undefined value.

Can the limit be evaluated using other methods?

Yes, there are other methods, such as L'Hôpital's rule, that can be used to evaluate limits of functions with singularities. However, in this case, the application of L'Hôpital's rule also results in an undefined value.

Is it possible for the limit to have a specific value?

No, the limit of (x + (1/x)) sin(x) as x approaches 0 is undefined, and therefore, it does not have a specific value. It is important to note that this does not mean that the function itself is undefined, just that the limit cannot be determined.

How can we interpret the undefined limit in terms of the graph of the function?

The undefined limit indicates that there is a discontinuity in the graph of the function at x = 0. This can be visualized as a "hole" or a "jump" in the graph. It is a point where the function is not continuous and cannot be extended to create a smooth curve.

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