Basic theorems on limits Assignment Help

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Basic theorems on limits:

Suppose 1187_Basic theorems on limits.pngf(x) = l1 and 1187_Basic theorems on limits.png g(x) = l2, where l1 and l2 are finite, then the subsequent theorems on limits may be used to calculate the limits

(i)         1187_Basic theorems on limits.png(c1 f(x) ± c2 g(x)) = c1 l1 ± c2 l2, where c1and c2 are provided constants.

(ii)        1187_Basic theorems on limits.pngf(x). g(x) =1187_Basic theorems on limits.png f (x).1187_Basic theorems on limits.pngg (x)  = l1. l2

(iii)       698_Basic theorems on limits2.png

(iv)      1187_Basic theorems on limits.png f (g(x)) = f (1187_Basic theorems on limits.pngg(x)) = f(l2), if and only if  f(x) is continuous at x = l2.

            For example649_Basic theorems on limits3.png (where [.]  shows the greatest integer function)

            Here [x] is not continuous at x = 1. Also 2121_Basic theorems on limits4.png

 (v)       If f(x) ≤ g(x) ∀ x ∈ R, then 1187_Basic theorems on limits.png f(x) ≤ 1187_Basic theorems on limits.png g(x).

Note: We need to be very cautious while applying these theorems. For example if we can try to use the theorems on 1536_Reason for non-existence of the limit2.pngsinx/x=1 we obtain 1536_Reason for non-existence of the limit2.png=  sin x/x. 1536_Reason for non-existence of the limit2.pngsinx/x, which could not exist.

Which is an strange result, because in that case the applied limit may not be given as the multiplication of two limits as 1536_Reason for non-existence of the limit2.png 1/x does not exist. 

 

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