MIT18_014F10_ex3_sols

# MIT18_014F10_ex3_sols - Exam 3 Solutions Problem 1 Evaluate...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exam 3 Solutions Problem 1. Evaluate lim x → x 1 − 1 . log( x +1) Solution We begin by writing the problem as a single fraction, lim x log( x +1) − x . → x log( x +1) Observe that both numerator and denominator have limit zero, and thus we can apply L’Hopital’s rule to see lim log( x + 1) − x = lim 1 / ( x + 1) − 1 = lim 1 − ( x + 1) . x → x log( x + 1) x → log( x + 1) + x/ ( x + 1) x → ( x + 1) log( x + 1) + x Note that in the expression on the right, the limits of both numerator and denominator are again zero. Thus a second application of L’Hopital’s rule gives 1 − ( x + 1) − 1 1 x lim ( x + 1) log( x + 1) + x = lim x → log( x + 1) + ( x + 1) / ( x + 1) + 1 = − 2 . → 3 x − 2 Problem 2. Evaluate x 2 − 6 x +10 dx. Solution We start by observing that the denominator can be written as ( x − 3) 2 + 1. That makes part of the problem easy: dx − 2 ( x − 3) 2 + 1 = − 2 arctan( x − 3) ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

MIT18_014F10_ex3_sols - Exam 3 Solutions Problem 1 Evaluate...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online