{"id":163494,"date":"2023-11-17T10:38:57","date_gmt":"2023-11-17T05:08:57","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=163494"},"modified":"2023-11-17T10:40:52","modified_gmt":"2023-11-17T05:10:52","slug":"ml-aggarwal-class-12-maths-solutions-section-a-chapter-6-ex-6-4","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-12-maths-solutions-section-a-chapter-6-ex-6-4\/","title":{"rendered":"ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.4"},"content":{"rendered":"

Well-structured ML Aggarwal Class 12 Solutions<\/a> Chapter 6 Indeterminate Forms Ex 6.4 facilitate a deeper understanding of mathematical principles.<\/p>\n

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.4<\/h2>\n

Typical Problems:<\/span><\/p>\n

Question 1.
\n(i) Evaluate the following limits:
\n(i) \\(\\underset{x \\rightarrow \\infty}{\\mathbf{L t}}\\) x tan-1<\/sup> (\\(\\frac{2}{x}\\))
\n(ii) \\(\\underset{x \\rightarrow \\infty}{\\mathbf{L t}}\\) \\(\\frac{\\sin ^{-1} \\frac{1}{x}}{\\tan \\frac{1}{x}}\\)
\nSolution:
\n(i) Put \\(\\frac{1}{x}\\) = t
\nas x \u2192 \u221e \u21d2 t \u2192 0
\n\u2234 \\(\\underset{x \\rightarrow \\infty}{\\mathbf{L t}}\\) x tan-1<\/sup> (\\(\\frac{2}{x}\\)) = \\(\\underset{t \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\tan ^{-1}(2 t)}{t}\\)
\n(\\(\\frac{0}{0}\\) form, using L’Hopital’s rule)
\n= \\(\\ {Lt}_{t \\rightarrow 0} \\frac{2}{1+4 t^2}\\)
\n= \\(\\frac{2}{1+4 \\times 0}\\) = 2.<\/p>\n

(ii) Put \\(\\frac{1}{x}\\) = t
\n\u21d2 x = \\(\\frac{1}{t}\\)
\nas x \u2192 \u221e \u21d2 t \u2192 0
\n\u2234 \\(\\ {Lt}_{x \\rightarrow \\infty} \\frac{\\sin ^{-1} \\frac{1}{x}}{\\tan \\frac{1}{x}}=\\underset{t \\rightarrow 0}{\\mathrm{Lt}} \\frac{\\sin ^{-1} t}{\\tan t}\\)
\n(\\(\\frac{0}{0}\\) form, using L’Hopital’s rule)
\n= \\(\\ {Lt}_{t \\rightarrow 0} \\frac{\\frac{1}{\\sqrt{1-t^2}}}{\\sec ^2 t}\\)
\n= \\(\\ {Lt}_{t \\rightarrow 0} \\frac{1}{\\sec ^2 t \\sqrt{1-t^2}}\\)
\n= \\(\\frac{1}{(1)^2 \\times \\sqrt{1-0^2}}\\) = 1.<\/p>\n

\"ML<\/p>\n

Question 2.
\nIf \\(\\frac{\\sin 2 x+k \\sin x}{x^3}\\) is finite, find k and the limit.
\nSolution:
\n\\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\sin 2 x+k \\sin x}{x^3}\\)
\n(\\(\\frac{0}{0}\\) form, using L’Hopital’s rule)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{2 \\cos 2 x+k \\cos x}{3 x^2}\\) …………(1)
\nNow deno 3x2<\/sup> \u2192 0 as x \u2192 0 and limit is given to be finite.
\nSo it is necessary that Numerator of eqn. (1)
\ni.e. 2 cos 2x + k cos x vanishes as x \u2192 0
\nand this happens when 2 + k = 0
\n\u21d2 k = – 2
\n\u2234 From (1) ; we have
\ngiven limit = \\(\\ {Lt}_{x \\rightarrow 0} \\frac{2 \\cos 2 x-2 \\cos x}{3 x^2}\\) ((\\(\\frac{0}{0}\\) form)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{-4 \\sin 2 x+2 \\sin x}{6 x}\\) ((\\(\\frac{0}{0}\\) form)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{-8 \\cos 2 x+2 \\cos x}{6}\\)
\n= \\(\\frac{-8 \\times 1+2 \\times 1}{6}\\) = – 1.<\/p>\n

Question 3.
\nFind the values of a and b \\(\\ {Lt}_{x \\rightarrow 0} \\frac{x(1-a \\cos x)+b \\sin x}{x^3}\\) exists and equals \\(\\frac{1}{3}\\).
\nSolution:
\n\\(\\ {Lt}_{x \\rightarrow 0} \\frac{x(1-a \\cos x)+b \\sin x}{x^3}\\)
\n(\\(\\frac{0}{0}\\) form, using L’Hopital’s rule)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{(1-a \\cos x)+a x \\sin x+b \\cos x}{3 x^2}\\) ………….(1)
\nNow the denominator of eqn. (1)
\ni.e. 3x2<\/sup> \u2192 0 as x \u2192 0.
\n\u2234 the given limit exists finitely, it is necessary that Numerator of eqn. (1) is also goes to 0 as x \u2192 0 and this happen when
\n1 – a \u00d7 1 + a \u00d7 0 + b = 0
\n\u21d2 1 – a + b = 0
\n\u21d2 a – b = 1 ………..(2)
\nLet eqn. (2) happens, Then given limit
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{(1-a \\cos x)+a x \\sin x+b \\cos x}{3 x^2}\\left(\\frac{0}{0}\\right)\\)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{a \\sin x+a \\sin x+a x \\cos x-b \\sin x}{6 x}\\)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{2 a \\sin x+a x \\cos x-b \\sin x}{6 x}\\left(\\frac{0}{0}\\right)\\)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{2 a \\cos x+a \\cos x-a x \\sin x-b \\cos x}{6}\\)
\n= \\(\\frac{2 a+a-b}{6}\\)
\nAlso given limit = \\(\\frac{1}{3}\\)
\n\u2234 \\(\\frac{2 a+a-b}{6}\\) = \\(\\frac{1}{3}\\)
\n\u21d2 3a – b = 2 ………..(3)
\nOn solving eqn. (2) and eqn. (3) ; we have
\na = \\(\\frac{1}{2}\\)
\nand b = – \\(\\frac{1}{2}\\).<\/p>\n

\"ML<\/p>\n

Question 4.
\nFind the values of a, b and c so that \\(\\ {Lt}_{x \\rightarrow 0} \\frac{a e^x-b \\cos x+c e^{-x}}{x \\sin x}\\) = 2.
\nSolution:
\n\\(\\ {Lt}_{x \\rightarrow 0} \\frac{a e^x-b \\cos x+c e^{-x}}{x \\sin x}\\)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{a e^x-b \\cos x+c e^{-x}}{x^2}\\left(\\frac{x}{\\sin x}\\right)\\)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{a e^x-b \\cos x+c e^{-x}}{x^2}\\) . 1 ……….(1)
\n[\u2235 \\(\\underset{\\theta \\rightarrow 0}{\\ {Lt}} \\frac{\\theta}{\\sin \\theta}\\) = 1]
\nHere denominator of eqn. (1) = x2<\/sup> \u2192 0 as x \u2192 0.
\nNow the limit exists finitely if Numerator of eqn. (1)
\ni.e. aex<\/sup> – b cos x + c e-x<\/sup> as x \u2192 0
\nand thos happen if a – b + c = 0 ……….(2)
\nsuppose eqn.(2) is holds. Then given limit
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{a e^x-b \\cos x+c e^{-x}}{x^2}\\) (\\(\\frac{0}{0}\\) form)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{a e^x+b \\sin x-c e^{-x}}{2 x}\\) ………..(3)
\nNow denominator of eqn. (3) i.e. 2x \u2192 0 as x \u2192 0
\n\u2234 in order that the limit exists finitely it is necessary that Numerator of eqn. (3) vanishes as x \u2192 0.
\nThis happens if a \u00d7 1 + b \u00d7 0 – c = 0
\n\u21d2 a – c = 0 ………..(4)
\nNow eqn. (4) holds.
\nThen given limit = \\(\\ {Lt}_{x \\rightarrow 0} \\frac{a e^x+b \\sin x-c e^{-x}}{2 x}\\left(\\frac{0}{0}\\right)\\)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{a e^x+b \\cos x+c e^{-x}}{2}\\)
\n= \\(\\frac{a+b+c}{2}\\)
\nAlso given limit = 2
\n\u2234 a + b + c = 4 ……..(5)
\nfrom (4) ;
\na = c
\n\u2234 from (1) ;
\n2c – b = 0 …………..(6)
\nand from (5) ;
\nb + 2c = 4 …………..(7)
\nOn solving eqn. (6) and (7) ; we have
\nc = 1; a = 1; b = 2.<\/p>\n

Question 5.
\nEvaluate the following limits :
\n(i) \\(\\underset{x \\rightarrow \\infty}{\\ {Lt}} \\frac{\\log _a x}{x^k}\\), k > 0
\n(ii) \\(\\ {Lt}_{x \\rightarrow \\infty} 2^x \\sin \\frac{a}{2^x}\\), a \u2260 0
\nSolution:
\n(i) \\(\\underset{x \\rightarrow \\infty}{\\ {Lt}} \\frac{\\log _a x}{x^k}\\), k > 0
\n(\\(\\frac{\\infty}{\\infty}\\) form, using L’Hopital’s rule)
\n= \\(\\ {Lt}_{x \\rightarrow \\infty} \\frac{\\frac{1}{x} \\frac{1}{\\log a}}{k x^{k-1}}\\)
\n= \\(\\ {Lt}_{x \\rightarrow \\infty} \\frac{1}{(\\log a) k x^k}\\), k > 0 = 0<\/p>\n

(ii) put \\(\\frac{a}{2^x}\\) = t
\nas x \u2192 \u221e
\n\u21d2 2x<\/sup> \u2192 \u221e
\n\u21d2 \\(\\frac{a}{2^x}\\) \u2192 0
\n\u21d2 t \u2192 0 [a \u2260 0]
\n\u2234 \\(\\underset{x \\rightarrow \\infty}{\\mathbf{L t}}\\) 2x<\/sup> sin \\(\\frac{a}{2^x}\\)
\n= \\(\\underset{t \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{a}{t}\\) sin t
\n= a \\(\\underset{x \\rightarrow \\infty}{\\mathbf{L t}}\\) \\(\\frac{sin t}{t}\\)
\n= a \u00d7 1 = a.<\/p>\n

\"ML<\/p>\n

Question 6.
\nEvaluate the following limits:
\n(i) \\(\\ {Lt}_{x \\rightarrow 0}\\left(\\frac{\\tan x}{x}\\right)^{1 \/ x}\\)
\n(ii) \\(\\ {Lt}_{x \\rightarrow a}\\left(2-\\frac{a}{x}\\right)^{\\tan \\frac{\\pi x}{2 a}}\\).
\nSolution:
\n(i) Let F(x) = \\(\\ {Lt}_{x \\rightarrow 0}\\left(\\frac{\\tan x}{x}\\right)^{1 \/ x}\\)
\n\u21d2 log F(x) = \\(\\frac{1}{x}\\) log \\(\\left(\\frac{\\tan x}{x}\\right)\\)<\/p>\n

\"ML<\/p>\n

(ii) Let F(x) = \\(\\left(2-\\frac{a}{x}\\right)^{\\tan \\frac{\\pi x}{2 a}}\\)<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 7.
\nEvaluate the following limits:
\n(i) \\(\\ {Lt}_{x \\rightarrow 0}\\left(\\frac{\\tan x}{x}\\right)^{1 \/ x^2}\\)
\n(ii) \\(\\ {Lt}_{x \\rightarrow 0}(1+2 x)^{\\frac{x+5}{x}}\\)
\nSolution:
\n(i) \\(\\ {Lt}_{x \\rightarrow 0}\\left(\\frac{\\tan x}{x}\\right)^{1 \/ x^2}\\)<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

\u2234 log (\\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) F(x)) = \\(\\frac{1}{3}\\)
\n\u21d2 F(x) = e1\/3<\/sup><\/p>\n

(ii) Let F(x) = \\(\\ {Lt}_{x \\rightarrow 0}(1+2 x)^{\\frac{x+5}{x}}\\)
\n\u21d2 log F(x) = \\(\\left(\\frac{x+5}{x}\\right)\\) log (1 + 2x)
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) log F(x) = \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\left(\\frac{x+5}{x}\\right)\\) log (1 + 2x)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{(x+5) \\log (1+2 x)}{x}\\) (\\(\\frac{0}{0}\\) form)
\n= \\(\\ {Lt}_{x \\rightarrow 0} \\frac{\\frac{(x+5) 2}{1+2 x}+\\log (1+2 x)}{1}\\)
\n= \\(\\frac{(0+5) 2}{1+0}\\) + log (1 + 0) = 0
\n\u2234 log \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) F(x) = 10
\n\u21d2 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\ {Lt}_{x \\rightarrow 0}(1+2 x)^{\\frac{x+5}{x}}\\) = e10<\/sup><\/p>\n

\"ML<\/p>\n

Question 8.
\nEvaluate : \\(\\ {Lt}_{x \\rightarrow 0}\\left(\\frac{\\sin x}{x}\\right)^{\\frac{\\sin x}{x-\\sin x}}\\)
\nSolution:
\n\\(\\ {Lt}_{x \\rightarrow 0}\\left(\\frac{\\sin x}{x}\\right)^{\\frac{\\sin x}{x-\\sin x}}\\) = \\(\\ {Lt}_{x \\rightarrow 0}\\left(\\frac{\\sin x}{x}\\right)^{\\frac{\\frac{\\sin x}{x}}{1-\\frac{\\sin x}{x}}}\\)<\/p>\n

\"ML<\/p>\n","protected":false},"excerpt":{"rendered":"

Well-structured ML Aggarwal Class 12 Solutions Chapter 6 Indeterminate Forms Ex 6.4 facilitate a deeper understanding of mathematical principles. ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.4 Typical Problems: Question 1. (i) Evaluate the following limits: (i) x tan-1 () (ii) Solution: (i) Put = t as x \u2192 …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/163494"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=163494"}],"version-history":[{"count":5,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/163494\/revisions"}],"predecessor-version":[{"id":163521,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/163494\/revisions\/163521"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=163494"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=163494"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=163494"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}