Can you solve for the following limit without using L'Hôpital's rule ?
x → 0 + lim x sin x
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You should let the limit approach 0 from above (as in x → 0 + ) , since x sin x is not defined for x < 0 .
Since sin x ≤ x , therefore sin x approaches zero earlier than x . So in the limit when x tends to zero, x sin x becomes x 0 when x is still not zero. Hence the required limit is 1 .
Taylor expansions are much more powerful and versatile than l'Hôpital's rule, use them !
x sin ( x ) = e ln ( x ) sin ( x ) = e ln ( x ) ( x + o ( x ) ) x → 0 ⟶ e 0 = 1
Note: x ln ( x ) x → 0 ⟶ 0
We can write x → 0 lim x sin x as x → 0 lim e sin x ⋅ ln x .
As ln ( x ) = − n = 1 ∑ ∞ n ( − 1 ) n ( x − 1 ) n This suggests, as you put x = 0 , ln x = c ∣ c < 0 is a constant. But sin ( 0 ) = 0 .
Thus, x → 0 lim e sin x ln x ⟹ e 0 = 1 which is the answer.
ln ( 0 ) = − ∞ ... Not c !
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I'll present the L'Hopital method: let L = x → 0 lim x sin x taking natural log on both sides, ln L = x → 0 lim sin x ln x = x → 0 lim csc x ln x = x → 0 lim d x d csc x d x d ln x = x → 0 lim − cot x csc x x 1 = x → 0 lim − x cos x sin 2 x = x → 0 lim − x sin x tan x = 0
Thus, L = e 0 = 1