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Chapter 3: Q. 46 (page 311)

Calculate each of the limits in Exercises 21-48. Some of these limits are made easier by L鈥橦opital鈥檚 rule, and some are not.

limx鈫�1sin(lnx)x-1.

Short Answer

Expert verified

The exact value of the limitlimx鈫�1sin(lnx)x-1is,1.

Step by step solution

01

Step 1. Given information

limx鈫�1sin(lnx)x-1.

02

Step 2. The limit using L'Hopital's rule is given below:

limx鈫�1sin(lnx)x-1=limx鈫�1sin(lnx)x-1[ in the form of 00; Using L'Hopital's rule]

=limx鈫�1cos(lnx)路1x1-0=limx鈫�1cos(lnx)路1x1=cos(ln1)路11=1.1=1

Therefore, the exact value is,1.

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Most popular questions from this chapter

Calculate each of the limits in Exercises 15鈥�20 (a) using

L鈥橦opital鈥檚 rule and (b) without using L鈥橦 藛 opital鈥檚 rule.

limx鈫�0cosx-1sinx

Sketch careful, labeled graphs of each function f in Exercises 63鈥�82 by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of f,f',andf'', and examine any relevant limits so that you can describe all key points and behaviors of f.

f(x)=x2-1x2-5x+4

Find the possibility graph of its derivative f'.

For each set of sign charts in Exercises 53鈥�62, sketch a possible graph of f.

Use the first derivative test to determine the local extrema of each function in Exercises 39- 50. Then verify your algebraic answers with graphs from a calculator or graphing utility.

f(x)=(x-1)2x+2
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