Can you squeeze the limit?

Calculus Level 1

Evaluate the following limit , lim ⁡ x → ∞ x + 7 sin ⁡ x − 2 x + 13 \lim_{x \to \infty} \frac{x+7\sin{x}}{-2x+13}

Try solving the problem using this

1 2 \frac12 − 1 2 -\frac12 8 9 \frac89 2 3 \frac23

Athiyaman Nallathambi by Athiyaman Nallathambi , 28, United Arab Em…

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3 solutions

Chew-Seong Cheong
Aug 16, 2015

lim ⁡ x → ∞ x + 7 sin ⁡ x − 2 x + 13 = lim ⁡ x → ∞ 1 + 7 sin ⁡ x x − 2 + 13 x = 1 + 0 − 2 + 0 = − 1 2 \begin{aligned} \lim_{x \to \infty} \frac{x+7\sin{x}}{-2x+13} & = \lim_{x \to \infty} \frac{1+\frac{7\sin{x}}{x}}{-2+\frac{13}{x}} \\ & = \frac {1+0} {-2+0} = \boxed{-\dfrac{1}{2}} \end{aligned}

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Moderator note:

Simple standard approach.

We know that, − 1 ≤ sin ⁡ x ≤ 1 -1\leq \sin{x} \leq 1 ⇒ ( − 7 ≤ 7 sin ⁡ x ≤ 7 ) ⇒ ( x − 7 ≤ x + 7 sin ⁡ x ≤ x + 7 ) \Rightarrow (-7 \leq 7\sin{x} \leq 7 )\Rightarrow (x-7 \leq x+7\sin{x} \leq x+7) Dividing by − 2 x + 13 -2x+13 we get, x − 7 − 2 x + 13 ≥ x + 7 sin ⁡ x − 2 x + 13 ≥ x + 7 − 2 x + 13 \frac{x-7}{-2x+13} \geq \frac{x+7\sin{x}}{-2x+13} \geq \frac{x+7}{-2x+13} By Squeeze Theorem , if, lim ⁡ x → ∞ x − 7 − 2 x + 13 = lim ⁡ x → ∞ x + 7 − 2 x + 13 = L \lim_{x \to \infty} \frac{x-7}{-2x+13} = \lim_{x \to \infty} \frac{x+7}{-2x+13} = L then, lim ⁡ x → ∞ x + 7 sin ⁡ x − 2 x + 13 = L \lim_{x \to \infty} \frac{x+7\sin{x}}{-2x+13} = L Evaluating limits we get, lim ⁡ x → ∞ x − 7 − 2 x + 13 = lim ⁡ x → ∞ x + 7 − 2 x + 13 = − 1 2 \lim_{x \to \infty} \frac{x-7}{-2x+13} = \lim_{x \to \infty} \frac{x+7}{-2x+13} = -\frac{1}{2} Therefore by Squeeze Theorem \text{Squeeze Theorem} , lim ⁡ x → ∞ x + 7 sin ⁡ x − 2 x + 13 = − 1 2 \lim_{x \to \infty} \frac{x+7\sin{x}}{-2x+13} = \boxed{-\dfrac{1}{2}}

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Moderator note:

It is not necessary to apply Squeeze Theorem. There's a much simpler approach.

Hint : For large x x , x + 7 sin ⁡ ( x ) ≈ x x + 7\sin(x) \approx x .

Nice use of sandwich theorem or squeeze theorem ,but you can also simply solve it using L-hospital .

Akhil Bansal - 5 years, 10 months ago

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Yeah well i had just learnt about the squeeze theorem and wanted to apply it in a question and .....I did. :)

Athiyaman Nallathambi - 5 years, 10 months ago

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Thanks for sharing multiple ways to approach a problem.

Calvin Lin Staff - 5 years, 10 months ago

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@Calvin Lin – Not a problem at all sir

Athiyaman Nallathambi - 5 years, 9 months ago

no need for the L-hospital Rule simple simplify and work

Ramez Hindi - 5 years, 9 months ago

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You can check out Chew-Seong Cheong sir's solutions , because that is how I would solve it if it appears in any time based exam or Olympiad.If your method is different please post it as a solution here as it would be awesome to see multiple ways to solve this problem

Athiyaman Nallathambi - 5 years, 9 months ago
Ashish Menon
Mar 7, 2016


Observing graph, we find that the ansee is − 1 2 -{\Large {\frac{1}{2}}} . □ _\square

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