Limits! I love' em

Calculus Level 3

lim ⁡ x → − ∞ x 4 sin ⁡ ( 1 x ) + x 2 1 + ∣ x ∣ 3 = ? \lim_{x\to-\infty } \dfrac{x^4 \sin\left( \tfrac1x\right) + x^2}{1+|x|^3} = \, ?


The answer is -1.

Arghyadeep Chatterjee by Arghyadeep Chatterjee , 22, India

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2 solutions

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The last step of applying L'Hospital you can do it by just observing that sin h / h will become 1 and in denominator it will be -1 . I just did it because I love to use L'Hospital.

Arghyadeep Chatterjee - 3 years, 1 month ago
N. Aadhaar Murty
Mar 8, 2021

We can substitute x = − t x=-t to get rid of the absolute value -

L = lim ⁡ x t o − ∞ x 4 sin ⁡ ( 1 x ) + x 2 1 + ∣ x ∣ 3 = lim ⁡ t → ∞ − t 4 sin ⁡ ( 1 t ) + t 2 1 + t 3 L = \lim_{x\ to -\infty} \frac {x^4 \sin(\frac {1}{x})+x^2}{1 + |x|^3} = \lim_{t \to \infty} \frac {-t^4 \sin(\frac {1}{t}) + t^2}{1 + t^3}

Since we get a ∞ ∞ \frac {\infty}{\infty} form, we can use L'Hopital -

L = lim ⁡ t → ∞ − 4 t 3 sin ⁡ ( 1 t ) + t 2 cos ⁡ ( 1 t ) + 2 t 3 t 2 = lim ⁡ t → ∞ − 4 3 t sin ⁡ ( 1 t ) + cos ⁡ ( 1 t ) 3 + 2 3 t = − 4 3 ⋅ 1 + 1 3 = − 1 L =\lim_{t \to \infty} \frac {-4t^3 \sin(\frac {1}{t}) + t^2 \cos(\frac {1}{t}) + 2t}{3t^2} = \lim_{t \to \infty} \frac {-4}{3}t \sin\left(\frac {1}{t}\right) + \frac {\cos(\frac {1}{t})}{3} + \frac {2}{3t} = \frac {-4}{3} \cdot 1 + \frac {1}{3} = \boxed {-1}

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