What is the limit as h approaches 0 of hsin(1/h) ?

The limit as h approaches 0 of hsin(1/h) is 0

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limit as h approaches 0 of hsin(1/h)

h \to 0 lim (h sin (\frac{1}{h}))

0

Solution steps

h \to 0 lim (h sin (\frac{1}{h}))

Apply the Squeeze Theorem:

0

= 0

\int \frac{1}{6} x^{2} d x

\int_{0}^{4} 8 e^{- 3 x} d x

f (x) = x^{3} - x + 1, at x = 3

\int_{1}^{7} \frac{2}{x ^{3}} d x

tan (\frac{x}{2})

What is the limit as h approaches 0 of hsin(1/h) ?
The limit as h approaches 0 of hsin(1/h) is 0