Evaluate the limit. lim (x² – x')e2*I -00

Answered: lim (x² – x')e2* I -00

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Evaluate the limit.

Expert Solution

Step 1

Given: $\lim_{x \to - \infty} (x^{2} - x^{3}) e^{2 x}$

for evaluating given limit, we substitute $x = - h$ , after that simplify it and then use L hospital rule

according to L hospital's rule

if, $\lim_{x \to \infty} f (x) \to \infty, \lim_{x \to \infty} g (x) \to \infty$

then

$\lim_{x \to \infty} \frac{f (x)}{g (x)} = \lim_{x \to \infty} \frac{f' (x)}{g' (x)}$

Step 2

so,

$\begin{array}{rcl} \lim_{x \to - \infty} (x^{2} - x^{3}) e^{2 x} & = & \lim_{- h \to - \infty} [{(- h)}^{2} - {(- h)}^{3}] e^{2 (- h)} \\ = & \lim_{h \to \infty} [h^{2} + h^{3}] e^{- 2 h} \\ = & \lim_{h \to \infty} \frac{h^{2} + h^{3}}{e^{2 h}} \end{array}$

now given limit has been converted in $(\frac{\infty}{\infty})$ form , so we will use L hospital rule

so,

$\lim_{h \to \infty} \frac{h^{2} + h^{3}}{e^{2 h}} = \lim_{h \to \infty} \frac{2 h + 3 h^{2}}{2 e^{2 h}} (∵ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (e^{k x}) = k e^{k x})$

again using L hospital's rule

$\begin{array}{rcl} \lim_{h \to \infty} \frac{2 h + 3 h^{2}}{2 e^{2 h}} & = & \lim_{h \to \infty} \frac{2 + 3 (2 h)}{2 (2 e^{2 h})} \\ = & \lim_{h \to \infty} \frac{0 + 6}{4 (2 e^{2 h})} \\ = & \lim_{h \to \infty} \frac{3}{4} (\frac{1}{e^{2 h}}) \\ = & \frac{3}{4} (\frac{1}{\infty}) \\ = & 0 \end{array}$

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