x → 0 lim ∣ x ∣ sin x
Find the above limit, if it exists.
Clarification : e ≈ 2 . 7 1 8 2 8 is the Euler's number .
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Simple standard approach.
The limit L = x → 0 lim ∣ x ∣ sin x exists if x → 0 + lim ∣ x ∣ sin x = x → 0 − lim ∣ x ∣ sin x . Now consider:
L + = x → 0 + lim ∣ x ∣ sin x = x → 0 + lim x sin x = x → 0 + lim e ln x sin x Expressing e x by a more convenient form exp ( x ) = x → 0 + lim exp ( ln x sin x ) Dividing up and down by sin x = x → 0 + lim exp ( csc x ln x ) Since ln x → − ∞ and csc x → ∞ as x → 0 + , we can use the L’H o ˆ pital’s rule. = x → 0 + lim exp ( − csc x cot x x 1 ) Differentiating up and down = x → 0 + lim exp ( x cos x sin x sin x ) = exp ( 1 ( 1 ) ( 0 ) ) = e 0 = 1
Now, let us look at:
L − = x → 0 − lim ∣ x ∣ sin x = x → 0 + lim ∣ − x ∣ − sin x = x → 0 + lim x − sin x = x → 0 + lim x sin x 1 = L 1 + = 1
Therefore, L = L + = L − = 1
Easier way is to plot a graph of the given limit.
See at zero, the limit is one.Problem Loading...
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y = x → 0 lim x sin x
ln y = x → 0 lim ( sin x ) ( ln x ) = x → 0 lim csc x ln x
Using L'Hopital's Rule :
⋯ = x → 0 lim − cot x csc x x 1 = − x → 0 lim x cos x sin 2 x = − ( x → 0 lim cos x 1 ) ( x → 0 lim x sin 2 x ) = − x → 0 lim x sin 2 x ⋯ = − x → 0 lim 1 2 sin x cos x = − x → 0 lim sin 2 x = 0
ln y = 0 ⇒ y = e 0 = 1