That is y 2x 7 1 2 find limx0 x2 esin x by using the

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Unformatted text preview: ope 2, and it passes the point (2, −3), by the pointslope formula, the tangent line equation is y + 3 = 2(x − 2). That is, y = 2x − 7. 1 2. Find limx→0 x2 esin( x ) by using the Squeeze Theorem. Since the range of sine is always between −1 and 1, 1 −1 ≤ sin( ) ≤ 1. x Now, since y = ex is an increasing function, this inequality implies that 1 e−1 ≤ esin( x ) ≤ e1 , and by multiplying x2 ≥ 0 to each term above, we get 1 x2 e−1 ≤ x2 esin( x ) ≤ x2 e1 . Hence, since lim x2 e−1 = 0 = lim x2 e1 , x→ 0 x→ 0 1 by the Squeeze Theorem, it follows that lim...
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This note was uploaded on 02/10/2014 for the course MATH 235 taught by Professor Lucas during the Fall '08 term at James Madison University.

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