How can I calculate the following difficult limit: limx→∞(x+sin⁡x)sin⁡1x ?

To find this limit, you're going to need to employ the following two well-known facts from the theory of limits:limθ→0sin⁡θθ=1andlimθ→∞sin⁡θθ=0The proofs of those two statements are here and here respectively.Notice that as x approaches infinity, 1x approaches 0. So, x→∞⇒1x→0. Also, don't forget that x is the same thing as 11x as long as x does not equal zero (it doesn't in our case here):limx→∞[(x+sin⁡x)⋅sin⁡1x]=limx→∞(x⋅sin⁡1x)+limx→∞(sin⁡x⋅sin⁡1x)=lim1x→0sin⁡1x1x+limx→∞(xx⋅sin⁡x⋅sin⁡1x)=1+limx→∞sin⁡xx⋅lim1x→0sin⁡1x1x=1+0⋅1=1

limx→∞(x+sin⁡x)sin⁡1x==limx→∞sin⁡1x1x+0=1

To find this limit, you're going to need to employ the following two well-known facts from the theory of limits:limθ→0sin⁡θθ=1andlimθ→∞sin⁡θθ=0The proofs of those two statements are here and here respectively.Notice that as x approaches infinity, 1x approaches 0. So, x→∞⇒1x→0. Also, don't forget that x is the same thing as 11x as long as x does not equal zero (it doesn't in our case here):limx→∞[(x+sin⁡x)⋅sin⁡1x]=limx→∞(x⋅sin⁡1x)+limx→∞(sin⁡x⋅sin⁡1x)=lim1x→0sin⁡1x1x+limx→∞(xx⋅sin⁡x⋅sin⁡1x)=1+limx→∞sin⁡xx⋅lim1x→0sin⁡1x1x=1+0⋅1=1

limx→∞(x+sin⁡x)sin⁡1x==limx→∞sin⁡1x1x+0=1

Find out the answer to "How can I calculate the following difficult limit:..."

Erik Cantu

Answered question

2022-03-27

How can I calculate the following difficult limit:

lim_{x \to \infty} (x + \sin x) \sin \frac{1}{x}

Answer & Explanation

Boehm98wy

Beginner2022-03-28Added 18 answers

To find this limit, you're going to need to employ the following two well-known facts from the theory of limits:

lim_{θ \to 0} \frac{\sin θ}{θ} = 1

and

lim_{θ \to \infty} \frac{\sin θ}{θ} = 0

The proofs of those two statements are here and here respectively.
Notice that as x approaches infinity,

\frac{1}{x}

approaches 0. So,

x \to \infty \Rightarrow \frac{1}{x} \to 0

. Also, don't forget that x is the same thing as

\frac{1}{\frac{1}{x}}

as long as x does not equal zero (it doesn't in our case here):

lim_{x \to \infty} [(x + \sin x) \cdot \sin \frac{1}{x}] = lim_{x \to \infty} (x \cdot \sin \frac{1}{x}) + lim_{x \to \infty} (\sin x \cdot \sin \frac{1}{x}) =

lim_{\frac{1}{x} \to 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}} + lim_{x \to \infty} (\frac{x}{x} \cdot \sin x \cdot \sin \frac{1}{x}) =

1 + lim_{x \to \infty} \frac{\sin x}{x} \cdot lim_{\frac{1}{x} \to 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}} =

1 + 0 \cdot 1 = 1

Ashton Conrad

Beginner2022-03-29Added 11 answers

lim_{x \to \infty} (x + \sin x) \sin \frac{1}{x} =

= lim_{x \to \infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} + 0 = 1

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