{"id":4174,"date":"2024-01-30T12:24:31","date_gmt":"2024-01-30T06:24:31","guid":{"rendered":"https:\/\/stage-wp.10minuteschool.com\/?p=4174"},"modified":"2024-11-05T15:06:27","modified_gmt":"2024-11-05T09:06:27","slug":"determining-value-of-the-limit","status":"publish","type":"post","link":"https:\/\/10minuteschool.com\/content\/determining-value-of-the-limit\/","title":{"rendered":"\u09b2\u09bf\u09ae\u09bf\u099f\u09c7\u09b0 \u09ae\u09be\u09a8 \u09a8\u09bf\u09b0\u09cd\u09a3\u09af\u09bc \u0993 \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8\u09a4\u09be (Determining the value of the limit and Continuity of function)"},"content":{"rendered":"

\u09b2\u09bf\u09ae\u09bf\u099f \u0995\u09be\u0995\u09c7 \u09ac\u09b2\u09c7?<\/span><\/h2>\n

\u0995\u09cb\u09a8\u09cb \u098f\u0995\u099f\u09bf \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u099a\u09b2\u0995 \u09af\u0996\u09a8 \u0995\u09cb\u09a8\u09cb \u09a8\u09bf\u09b0\u09cd\u09a6\u09bf\u09b7\u09cd\u099f \u09ae\u09be\u09a8\u09c7\u09b0 \u0995\u09be\u099b\u09c7 \u0985\u0997\u09cd\u09b0\u09b8\u09b0 \u09b9\u09af\u09bc, \u09a4\u0996\u09a8 \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u09ae\u09be\u09a8 \u0995\u09cb\u09a8\u09cb \u09a8\u09bf\u09b0\u09cd\u09a6\u09bf\u09b7\u09cd\u099f \u09ae\u09be\u09a8\u09c7\u09b0 \u0995\u09be\u099b\u09c7 \u0985\u0997\u09cd\u09b0\u09b8\u09b0 \u09b9\u09b2\u09c7 \u09b8\u09c7\u0987 \u09a8\u09bf\u09b0\u09cd\u09a6\u09bf\u09b7\u09cd\u099f \u09ae\u09be\u09a8\u0995\u09c7\u0987 \u0986\u09ae\u09b0\u09be \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u09b8\u09c7\u0987 \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 \u09b2\u09bf\u09ae\u09bf\u099f<\/strong> \u09ac\u09b2\u09bf\u0964<\/p>\n

\u09b2\u09bf\u09ae\u09bf\u099f\u09c7\u09b0 \u09ae\u09be\u09a8 \u09a8\u09bf\u09b0\u09cd\u09a3\u09af\u09bc (Determining the value of the limit)<\/span><\/h2>\n

<\/h3>\n

Type-01 : <\/b>\u09b2\u09bf\u09ae\u09bf\u099f\u09c7\u09b0 \u0985\u09b8\u09cd\u09a4\u09bf\u09a4\u09b6\u09c0\u09b2\u09a4\u09be \u0995\u09c7\u09a8\u09cd\u09a6\u09cd\u09b0\u09bf\u0995\u00a0<\/strong><\/span><\/h3>\n

\u27f9Concept : <\/strong>\u0995\u09cb\u09a8<\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7<\/span>\u00a0\u09b2\u09bf\u09ae\u09bf\u099f<\/span><\/a><\/span> \u0985\u09b8\u09cd\u09a4\u09bf\u09a4\u09cd\u09a4\u09b6\u09c0\u09b2<\/span> \u09b9\u09ac\u09c7<\/span> \u09af\u09a6\u09bf<\/span> \u0990<\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7<\/span>\u00a0Left Hand Limit (LHL) = <\/span>Right Hand Limit (RHL) <\/span>\u09b9\u09df<\/span>\u0964<\/span> \u0985\u09b0\u09cd\u09a5\u09be\u09ce \\frac{\\lim }{x \\rightarrow 0^{-}} <\/span>f(x) <\/span>=\\frac{\\lim }{x \\rightarrow 0^{+}} <\/span>f(x) <\/span><\/span><\/p>\n

Example-01: \\frac{\\lim }{x \\rightarrow 0^{+}} \\frac{x}{\\sqrt{1-\\cos x}}=<\/span> ?<\/strong><\/p>\n\\operatorname{Sol}^{n}:<\/span>\n

 <\/p>\n

Right hand limit =\\frac{\\lim }{x \\rightarrow 0^{+}} \\frac{x}{\\sqrt{1-\\cos x}}=\\frac{\\lim }{x \\rightarrow 0^{+}} \\frac{x}{\\sqrt{2} \\sin \\frac{x}{2}}=\\frac{\\lim }{h \\rightarrow 0} \\frac{h}{\\sqrt{2} \\sin \\frac{h}{2}}[<\/span> Let, \\mathrm{x}=0=\\mathrm{h}]=\\sqrt{2}<\/span><\/p>\n

Left hand limit =\\frac{\\lim }{x \\rightarrow 0^{-}} \\frac{x}{\\sqrt{1-\\cos x}}=\\frac{\\lim }{h \\rightarrow 0} \\frac{0-h}{\\sqrt{1-\\cosh }}=\\frac{\\lim }{h \\rightarrow 0} \\frac{-h}{\\sqrt{2} \\sin \\frac{h}{2}}=-\\sqrt{2}<\/span><\/p>\n

\\because \\mathrm{RHL} \\neq \\mathrm{LHL}<\/span><\/span><\/p>\n

\\therefore<\/span> <\/span>\u09b2\u09bf\u09ae\u09bf\u099f\u09c7\u09b0<\/span> \u09ae\u09be\u09a8<\/span> \u0985\u09b8\u09cd\u09a4\u09bf\u09a4\u09cd\u09a4\u09b6\u09c0\u09b2<\/span>\u00a0 <\/span>\u09a8\u09df<\/span>\u0964<\/span><\/p>\n

Example-02: =\\frac{\\lim }{x \\rightarrow 0} \\frac{e^{|x|}-1}{x} ?<\/span><\/b><\/p>\n\\operatorname{Sol}^{n}:<\/span>\n

L. H. L. \\frac{\\lim }{x \\rightarrow 0^{-}} \\frac{e^{|x|}-1}{x}=-1 ;<\/span><\/p>\n

R.H.L. \\frac{\\lim }{x \\rightarrow 0^{+}} \\frac{e^{|x|}-1}{x}=1<\/span><\/p>\n

\\because \\mathrm{LHL} \\neq \\mathrm{RHL}<\/span><\/span><\/p>\n

\\therefore<\/span> limit <\/span>\u0985\u09b8\u09cd\u09a4\u09bf\u09a4\u09cd\u09a4\u09b6\u09c0\u09b2<\/span> \u09a8\u09df<\/span>\u0964<\/span>\u00a0 \u00a0 (Ans)<\/span><\/p>\n

<\/h3>\n

Type-02 : <\/b>\u09b9\u09b0\u09c7 \u09ac\u09b0\u09cd\u0997\u09ae\u09c2\u09b2 \u09b8\u0982\u09ac\u09b2\u09bf\u09a4 \u09aa\u09a6\u099f\u09bf \u0985\u09a8\u09c1\u09ac\u09a8\u09cd\u09a7\u09c0 \u09a6\u09bf\u09df\u09c7 \u0997\u09c1\u09a3 \u0995\u09b0\u09c7 \u09b2\u09bf\u09ae\u09bf\u099f \u09a8\u09bf\u09b0\u09cd\u09a3\u09df<\/strong>\u00a0<\/span><\/span><\/h3>\n

\u27f9 Concept:<\/strong> (i) \\frac{0}{0} \\text { form } <\/span> <\/span>(ii) \u09ac\u09c0\u099c\u0997\u09be\u09a3\u09bf\u09a4\u09bf\u0995 function (iii) \u09b2\u09ac \u09ac\u09be \u09b9\u09b0 \u0989\u09ad\u09df\u099f\u09bf\u09a4\u09c7 \u00a0 \\text { S } \\quad \\text { rt }(\\sqrt{ }) <\/span> <\/span>\u09a5\u09be\u0995\u09ac\u09c7 <\/span>\u0964<\/span><\/p>\n

\u09a8\u09bf\u09df\u09ae\u0983 \u0985\u09a8\u09c1\u09ac\u09a8\u09cd\u09a7\u09c0\u0995\u09b0\u09a8 \u0995\u09b0\u09c7 \u098f\u09ae\u09a8 \u09ac\u09be\u09a8\u09be\u09a4\u09c7 \u09b9\u09ac\u09c7 \u09af\u09c7\u09a8 limit<\/a><\/span> \u09ac\u09b8\u09be\u09a8\u09cb \u09af\u09be\u09df<\/span>\u0964<\/span><\/p>\n

Example-08:\u09b8\u09c0\u09ae\u09be\u09b8\u09cd\u09a5 \u09ae\u09be\u09a8 \u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0\u00a0 : \\frac{L t}{x \\rightarrow 0} <\/span>\\frac{2\\left(b-\\sqrt{\\left.b^{2}+x^{2}\\right)}\\right.}{x^{2}} <\/span><\/strong><\/p>\n

\\operatorname{Sol}^n <\/span>:<\/p>\n

\\frac{L t}{x \\rightarrow 0} <\/span>\\frac{2\\left(b-\\sqrt{\\left.b^{2}+x^{2}\\right)}\\right.}{x^{2}} <\/span>= <\/span>=\\frac{L t}{x \\rightarrow 0} <\/span>\\frac{2\\left(b^{2}-b^{2}-x^{2}\\right)}{x^{2}\\left(b+\\sqrt{\\left.b^{2}+x^{2}\\right)}\\right.} <\/span>=\\frac{-2}{\\left(b+\\sqrt{\\left.b^{2}+0\\right)}\\right.} <\/span>=-\\frac{1}{b} <\/span><\/p>\n

Example-09: \\frac{L t}{x \\rightarrow 0} <\/span>\\frac{\\sqrt{1-4 x}-\\sqrt{1-5 x}}{x} <\/span>=\\frac{L t}{x \\rightarrow 0} <\/span> \\frac{x}{x(\\sqrt{1-4 x}+(\\sqrt{1-5 x}))}<\/span>=\\frac{1}{2} <\/span>\u00a0 <\/b>(Ans.)<\/span><\/p>\n

[ <\/span>\u09b9\u09b0<\/span> \u0993<\/span> \u09b2\u09ac\u0995\u09c7<\/span> (\\sqrt{1-4 x} <\/span>+(\\sqrt{1-5 x}) <\/span>)\u09a6\u09cd\u09ac\u09be\u09b0\u09be<\/span> \u0997\u09c1\u09a3<\/span> \u0995\u09b0\u09c7<\/span> ]<\/span><\/p>\n

\u27f9 Limit Transfer :<\/span><\/p>\n

 <\/p>\n

Q. \\lim _{x \\rightarrow \\frac{\\pi}{2}} <\/span>\\frac{1-\\sin x}{\\cos x} <\/span><\/p>\n

\u09a7\u09b0\u09bf<\/span>, x=\\frac{\\pi}{2}+R <\/span><\/span> ; x \\rightarrow \\frac{\\pi}{2} <\/span>\u09b9\u09b2\u09c7 h \\rightarrow 0 <\/span><\/span>\u00a0<\/span>\u09b9\u09ac\u09c7<\/span>\u0964<\/span><\/p>\n

\u098f\u0996\u09a8,<\/span><\/p>\n

\\frac{\\lim }{x \\rightarrow \\frac{\\pi}{2}} <\/span>\\frac{1-\\sin x}{\\cos x} <\/span>=\\frac{1-\\sin x}{h \\rightarrow 0} \\frac{1-\\sin \\left(\\frac{\\pi}{2}+h\\right)}{\\cos \\left(\\frac{\\pi}{2}+h\\right)} <\/span><\/p>\n

=\\frac{\\lim }{h \\rightarrow 0} <\/span>\\frac{1-\\cos h}{-\\sin h} <\/span><\/p>\n

=\\frac{-\\lim }{h \\rightarrow 0}\u00a0 <\/span>\\frac{2 \\sin ^{2} \\frac{h}{2}}{\\sin h} <\/span><\/p>\n

=\\frac{-\\lim }{h \\rightarrow 0} <\/span>\\left(\\frac{\\sin \\frac{h}{2}}{\\frac{h}{2}}\\right)^{2} <\/span>x\\frac{h^{2}}{4} <\/span>x\\frac{\\lim }{h \\rightarrow 0} <\/span>\\frac{h}{\\sin h} <\/span>x\\frac{2}{h} <\/span><\/p>\n=1-x \\frac{\\lim }{h \\rightarrow 0} \\frac{h^{2}}{4} \\times \\frac{2}{h}=0 <\/span>\n

\\frac{\\lim }{x \\rightarrow \\frac{\\pi}{2}}\u00a0 <\/span>\\frac{1-\\sin x}{\\cos x} <\/span>\u00a0 (Ans)<\/span><\/p>\n

 <\/p>\n

<\/b>Q. \\frac{\\lim }{x \\rightarrow \\infty} 2^{x} \\sin \\frac{b}{2^{x}}<\/span>\u00a0<\/span><\/p>\n

\u27f9 <\/span>\u09a7\u09b0\u09bf<\/span>, \\frac{b}{2^{x}}=\\theta <\/span><\/span><\/p>\n

\\Rightarrow \\frac{b}{\\theta}=2^{x} <\/span><\/span><\/p>\n

 <\/p>\n

\u098f\u0996\u09a8,<\/span><\/p>\n

x \\rightarrow \\alpha <\/span> <\/span>\u09b9\u09b2\u09c7<\/span> 2^{x} \\rightarrow \\alpha \\therefore \\frac{b}{2^{x}} \\rightarrow 0 <\/span><\/p>\n\\begin{array}{l}\n\n\\therefore \\theta \\rightarrow 0 \\\\\n\n\\therefore \\frac{\\lim }{x \\rightarrow \\infty} 2^{x} \\sin \\frac{b}{2^{x}}=\\frac{\\lim }{\\theta \\rightarrow 0} \\frac{b}{\\theta} \\cdot \\sin \\theta \\\\\n\n=\\frac{\\lim }{\\theta \\rightarrow 0} \\frac{\\sin \\theta}{\\theta} \\times \\frac{\\lim }{\\theta \\rightarrow 0} \\theta \\times \\mathrm{b} \/ \\theta=6 \\\\\n\n\\text { (Ans) }\n\n\\end{array}<\/span>\n

 <\/p>\n

\u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8\u09a4\u09be <\/b>(Continuity of function)<\/b><\/span><\/h2>\n

\"\u09b2\u09bf\u09ae\u09bf\u099f<\/p>\n

\u27f9 \u0995\u09cb\u09a8\u09cb \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u09b2\u09c7\u0996\u099a\u09bf\u09a4\u09cd\u09b0\u09c7 \u0995\u09a5\u09be\u0993 \u099b\u09c7\u09a6 \u0995 \u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8\u09a4\u09be \u09a8\u09be \u09a5\u09be\u0995\u09b2\u09c7 \u09ab\u09be\u0982\u09b6\u09a8\u099f\u09bf\u0995\u09c7 \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8 \u09ab\u09be\u0982\u09b6\u09a8 \u09ac\u09b2\u09be \u09b9\u09df\u0964 \u0985\u09b0\u09cd\u09a5\u09be\u09ce x=a <\/span> \u09ac\u09bf\u09a8\u09cd\u09a7\u09c1\u09a4\u09c7 f(x) <\/span>\u09ab\u09be\u0982\u09b6\u09cd\u09a8\u0995\u09c7 \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8 \u09ac\u09b2\u09be \u09b9\u09df\u00a0 \u09af\u09a6\u09bf \\frac{\\lim }{x \\rightarrow a} <\/span> f(x) <\/span> \u09ac\u09bf\u09a6\u09cd\u09af\u09ae\u09be\u09a8 \u09a5\u09be\u0995\u09c7 \u098f\u09ac\u0982 \u09a4\u09be \u09b8\u09b8\u09c0\u09ae \u0993 f(a) <\/span><\/strong> \u098f\u09b0 \u09ae\u09be\u09a8 \u09b9\u09df\u0964<\/strong><\/p>\n

\u2234 <\/span>\u09ab\u09be\u0982\u09b6\u09a8<\/span> \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8<\/span> \u09b9\u09ac\u09c7<\/span> \u09af\u0996\u09a8<\/span> \\frac{\\lim }{x \\rightarrow a^{+}} f(x)=\\frac{\\lim }{x \\rightarrow a^{-}} f(x) <\/span><\/p>\n

 <\/p>\n

Q. f(x)=\\left\\{\\begin{array}{c}\n\n2 x+3 ;-\\frac{3}{2} \\leq x<0 \\\\\n\n-2 x+3 ; 0 \\leq x<\\frac{3}{2} \\\\\n\n-2 x-3 ; x \\geq \\frac{3}{2}\n\n\\end{array}\\right. <\/span><\/p>\n

x <\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 f(x) <\/span><\/span>\u00a0<\/span> \u098f\u09b0 \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8\u09a4\u09be \u09af\u09be\u099a\u09be\u0987 \u0995\u09b0<\/span>\u0964<\/span><\/p>\n

\u27f9 x=0 <\/span><\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 <\/span>–<\/p>\n \\frac{\\lim }{x \\rightarrow 0^{+}} f(x)=\\frac{\\lim }{x \\rightarrow 0^{+}}-2 x+3=3 <\/span>\n \\frac{\\lim }{x \\rightarrow 0^{-}} f(x)=\\frac{\\lim }{x \\rightarrow 0^{-}} 2 x+3=3 <\/span>\n

\u2234 \\therefore \\frac{\\lim }{x \\rightarrow 0^{-}} f(x)=\\frac{\\lim }{x \\rightarrow 0^{-}} f(x) <\/span><\/span><\/p>\n

x=0 <\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 <\/span>\u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8\u0964 (Ans)\u00a0<\/span><\/p>\n

Example-04: f(x)=\\left\\{\\begin{array}{c}\n\n2 x+3 ;-\\frac{3}{2} \\leq x<0 \\\\\n\n-2 x+3 ; 0 \\leq x<\\frac{3}{2} \\\\\n\n-2 x-3 ; x \\geq \\frac{3}{2}\n\n\\end{array}\\right. <\/span><\/b> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7<\/strong>\u00a0\u09ab\u09be\u0982\u09b6\u09be\u09a8\u099f\u09bf\u09b0 <\/b>\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8\u09a4\u09be<\/span>, <\/span>\u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8\u09a4\u09be<\/span> \u0986\u09b2\u09cb\u099a\u09a8\u09be<\/span> \u0995\u09b0<\/span>\u0964<\/span><\/p>\n

\\operatorname{Sol}^n <\/span> :<\/span><\/p>\n

x=0 <\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 <\/span>–<\/p>\n f(0)=0<\/span>\n \\frac{\\lim }{x \\rightarrow 0^{-}} f(x)=\\frac{\\lim }{h \\rightarrow 0} f(0-h)=\\frac{\\lim }{h \\rightarrow 0}(0-h)=0 <\/span>\n \\frac{\\lim }{x \\rightarrow 0^{+}} f(x)=\\frac{\\lim }{h \\rightarrow 0} f(0+h)=\\frac{\\lim }{h \\rightarrow 0}(0+h)=0 <\/span>\n

\u09af\u09c7\u09b9\u09c7\u09a4\u09c1 f(0)=\\frac{\\lim }{x \\rightarrow 0^{+}} f(x)=\\frac{\\lim }{x \\rightarrow 0^{-}} \\mathrm{f}(\\mathrm{x}) <\/span><\/span>\u00a0<\/span><\/p>\n

\u09b8\u09c1\u09a4\u09b0\u09be\u0982, x=0 <\/span><\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 \u09ab\u09be\u0982\u09b6\u09be\u09a8\u099f\u09bf \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8<\/span>\u0964<\/span><\/p>\n

x=1 <\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 –<\/span><\/p>\n

f(0)=1 <\/span> <\/span><\/p>\n

\\frac{\\lim }{x \\rightarrow 1^{-}} f(x)=\\frac{\\lim }{h \\rightarrow 0} f(1-h)=\\frac{\\lim }{h \\rightarrow 0}(1-h)=1 <\/span> <\/span><\/p>\n

\\frac{\\lim }{x \\rightarrow 1^{+}} f(x)=\\frac{\\lim }{h \\rightarrow 0} f(1+h)=\\frac{\\lim }{h \\rightarrow 0}\\{2-(1+h)\\}=1 <\/span> <\/span><\/p>\n

\u2235 f(0)=1 <\/span><\/span>=L.H.L=R.H.L<\/span><\/p>\n

\u2234 <\/span>\u09ab\u09be\u0982\u09b6\u09be\u09a8\u099f\u09bf<\/span> x=1 <\/span>\u00a0<\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8<\/span>\u0964<\/span><\/p>\n

x=2 <\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 –<\/span><\/p>\n

f(2)=2-2=0 <\/span> <\/span><\/p>\n \\frac{\\lim }{x \\rightarrow 2^{-}} f(x)=\\frac{\\lim }{h \\rightarrow 0} f(2+h)=\\frac{\\lim }{h \\rightarrow 0}[2-(2-h)]=0 <\/span>\n \\frac{\\lim }{x \\rightarrow 2^{+}} f(x)=\\frac{\\lim }{h \\rightarrow 0} f(2+h)=2 <\/span>\n

\\because f(2)=\\frac{\\lim }{x \\rightarrow 2^{-}} f(x) \\neq \\frac{\\lim }{x \\rightarrow 2^{+}} f(x) <\/span> <\/span><\/p>\n

\u2234<\/span> x=2 <\/span> \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7<\/span>\u00a0\u09ab\u09be\u0982\u09b6\u09be\u09a8\u099f\u09bf<\/span> \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8<\/span>\u0964<\/span><\/p>\n

\u2217 \u0997\u09cd\u09b0\u09be\u09ab \u0986\u0981\u0995\u09a4\u09c7 \u09aa\u09be\u09b0\u09b2\u09c7 \u0986\u09b0\u0993 \u09b8\u09b9\u099c\u09c7 \u09ac\u09c1\u099d\u09be \u09af\u09be\u09df \u0989\u09aa\u09b0\u09bf\u0989\u0995\u09cd\u09a4 \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u0997\u09cd\u09b0\u09be\u09ab\u0983<\/strong><\/p>\n

\"\u09b2\u09bf\u09ae\u09bf\u099f<\/p>\n

\u2217 \u09b8\u09cd\u099f\u09cd\u09af\u09be\u09a8\u09cd\u09a1\u0989\u0987\u099a\/\u09b8\u09cd\u0995\u09c1\u0987\u099c\/\u09aa\u09bf\u09a8\u099a\u09bf\u0982 \u0989\u09aa\u09aa\u09be\u09a6\u09cd\u09af (Sandwich \/ Squeeze \/ Pinching Theorem) :<\/strong><\/p>\n

\u099c\u09be\u09ae\u09be\u09b2 \u0995\u09be\u09ae\u09be\u09b2 \u0985\u09aa\u09c7\u0995\u09cd\u09b7\u09be\u09df \u09ac\u09c7\u09b6\u09bf \u0996\u09be\u09df, \u098f\u09ac\u0982 \u09a6\u09be\u09ae\u09be\u09b2 \u0985\u09aa\u09c7\u0995\u09cd\u09b7\u09be\u09df \u0995\u09bf\u09ae \u0996\u09be\u09df<\/span>\u0964<\/span> \u0995\u09cb\u09a8\u09cb \u098f\u0995\u09a6\u09bf\u09a8 \u0995\u09be\u09ae\u09be\u09b2 =2kg <\/span> \u098f\u09ac\u0982 \u09a6\u09be\u09ae\u09be\u09b2 =2kg <\/span> \u0996\u09be\u09df<\/span>\u0964<\/span> \u09a4\u09ac\u09c7 \u0995\u09be\u09ae\u09be\u09b2 \u0995\u09a4\u099f\u09c1\u0995\u09c1 \u0996\u09be\u09df?<\/span><\/p>\n

\u27f9 <\/span>\u0995\u09be\u09ae\u09be\u09b2<\/span> {K(x)} <\/span>\u00a0\u00a0<\/span><<\/span>\u099c\u09be\u09ae\u09be\u09b2<\/span> {J(x)} <\/span>\u00a0< <\/span>\u09a6\u09be\u09ae\u09be\u09b2<\/span> {D(x)} <\/span><\/p>\n

\u27f9 <\/span> K(x)<J(x)<D(x) <\/span><\/span><\/p>\n

\u27f9<\/span> 2<J(x)<2 <\/span><\/span><\/p>\n

\u2234 J(x)=2 <\/span><\/span><\/p>\n

\u09af\u09c7\u09ae\u09a8-<\/span><\/p>\n

1<\\frac{\\lim _{x \\rightarrow 0}}{x} \\frac{\\sin x}{x}<1 <\/span> <\/span><\/p>\n

\u2234 \\frac{\\lim }{x \\rightarrow 0} \\frac{\\sin x}{x}=1 <\/span><\/span>\u00a0<\/span><\/p>\n

 <\/p>\n \\text { Q. } \\frac{\\lim }{x \\rightarrow 0} x^{2} \\sin \\frac{1}{x}=?\u00a0<\/span>\n

\u27f9 <\/span>\u0986\u09ae\u09b0\u09be<\/span> \u099c\u09be\u09a8\u09bf<\/span>,<\/span><\/p>\n

\u27f9<\/span> -1 \\leq \\sin \\frac{1}{x} \\leq 1 <\/span>\u00a0<\/span><\/p>\n

\u27f9<\/span> -x^{2} \\leq x^{2} \\cdot \\sin \\frac{1}{x} \\leq x^{2} <\/span><\/span>\u00a0<\/span><\/p>\n

\u27f9<\/span> \\frac{\\lim }{x \\rightarrow 0}-x^{2} \\leq \\frac{\\lim }{x \\rightarrow 0} x^{2} \\cdot \\sin \\frac{1}{x} \\leq \\frac{\\lim }{x \\rightarrow 0} x^{2} <\/span><\/p>\n 0 \\leq \\frac{\\lim }{x \\rightarrow 0} x^{2} \\cdot \\sin \\frac{1}{x} \\leq 0 <\/span>\n

\\therefore \\frac{\\lim }{x \\rightarrow 0} x^{2} \\cdot \\sin \\frac{1}{x}=0<\/span>\u00a0 (Ans)<\/span><\/p>\n

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\u098f\u0987\u099a\u098f\u09b8\u09b8\u09bf \u0993 \u098f\u09a1\u09ae\u09bf\u09b6\u09a8 \u09aa\u09b0\u09c0\u0995\u09cd\u09b7\u09be\u09b0\u09cd\u09a5\u09c0\u09a6\u09c7\u09b0 \u099c\u09a8\u09cd\u09af \u0986\u09ae\u09be\u09a6\u09c7\u09b0 \u0995\u09cb\u09b0\u09cd\u09b8\u09b8\u09ae\u09c2\u09b9\u0983<\/strong><\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n