{"id":17643,"date":"2016-09-03T17:05:23","date_gmt":"2016-09-03T17:05:23","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=17643"},"modified":"2016-09-03T17:07:37","modified_gmt":"2016-09-03T17:07:37","slug":"calculus-why-limx%e2%86%920-sinxx-1","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/calculus-why-limx%e2%86%920-sinxx-1\/","title":{"rendered":"Calculus: why limx\u21920<\/sub> sinx\/x = 1"},"content":{"rendered":"

The tutor shows a reason for the famous limit x\u21920<\/sub> sinx\/x = 1.<\/h1>\n

<\/p>\n

The area of a sector of a circle about angle \u019f is \u019fr2<\/sup>\/2. (A sector is a pie slice.) The area of a triangle is base(height)\/2.<\/p>\n

In the above drawing, the angle is x. The radius of the circle is 1. The area of the inner sector is x(cosx)2<\/sup>\/2. The area of the triangle is cosx(sinx)\/2. The area of the outer sector is x(1)2<\/sup>\/2 = x\/2.<\/p>\n

Visually, the area of the inner sector is less than (or equal) that of the triangle, which in turn is less than (or equal) that of the outer sector. Therefore,<\/p>\n

x(cosx)2<\/sup>\/2 ≤ cosx(sinx)\/2 ≤ x\/2<\/p>\n

Multiplying by 2 gives<\/p>\n

x(cosx)2<\/sup> ≤ cosx(sinx) ≤ x<\/p>\n

Now, dividing by x gives<\/p>\n

(cosx)2<\/sup> ≤ cosx(sinx)\/x ≤ x\/x<\/p>\n

which simplifies to<\/p>\n

(cosx)2<\/sup> ≤ cosx(sinx)\/x ≤ 1<\/p>\n

Now, let’s imagine<\/p>\n

limx\u21920<\/sub>(cosx)2<\/sup> ≤ limx\u21920<\/sub>cosx(sinx)\/x ≤ limx\u21920<\/sub>1<\/p>\n

Cos(0)=1; of course, limx\u21920<\/sub>1 = 1. Hence, we have<\/p>\n

1 ≤ limx\u21920<\/sub>(sinx)\/x ≤ 1<\/p>\n

which means<\/p>\n

limx\u21920<\/sub>(sinx)\/x = 1 (Squeeze Theorem).<\/p>\n

HTH:)<\/p>\n

Source:<\/p>\n

Larson, Roland and Robert Hostetler. Calculus<\/u>, 3rd ed. Toronto: D C Heath and Company, 1989.<\/p>\n

Jack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"

The tutor shows a reason for the famous limit x\u21920 sinx\/x = 1. The area of a sector of a circle about angle \u019f is \u019fr2\/2. (A sector is a pie slice.) The area of a triangle is base(height)\/2. In …<\/p>\n

Calculus: why limx\u21920<\/sub> sinx\/x = 1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[234,3],"tags":[1861,1862],"class_list":["post-17643","post","type-post","status-publish","format-standard","hentry","category-calculus","category-math","tag-proof-lim-x0-sinxx-1","tag-why-lim-x0-sinxx-1"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17643","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=17643"}],"version-history":[{"count":24,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17643\/revisions"}],"predecessor-version":[{"id":17667,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17643\/revisions\/17667"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=17643"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=17643"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=17643"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}