Question:

<p><strong>1</strong>: Let t = x -1. As x \to 1, t \to 0.</p><p><strong>2</strong>: \lim_{t \to 0</p>} et-(1+t) tn .<p><strong>3</strong>: Expansion: et = 1 + t + t2</p> 2 + t3 6 + . . .<p><strong>4</strong>: Limit = \lim_{t \to 0</p>} t2 2 + t3 6 +... tn .<p><strong>5</strong>: For the limit to exist, the power in the denominator must be \leqthe lowest power in the</p> numerator. Thus n \leq2.