{"id":3358,"date":"2025-06-07T04:34:30","date_gmt":"2025-06-07T04:34:30","guid":{"rendered":"https:\/\/diznr.com\/?p=3358"},"modified":"2025-06-07T04:34:30","modified_gmt":"2025-06-07T04:34:30","slug":"cseit-gate-1996-subject-engineering-mathematics-topic-calulus-the-formula-used-compu-to","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/pdf\/cseit-gate-1996-subject-engineering-mathematics-topic-calulus-the-formula-used-compu-to\/","title":{"rendered":"CSEIT – GATE 1996\/ Subject – Engineering Mathematics\/ Topic – Calulus The formula used to compu."},"content":{"rendered":"

CSEIT – GATE 1996\/ Subject – Engineering Mathematics\/ Topic – Calulus The formula used to compu.<\/p>\n

[fvplayer id=”368″]<\/p>\n

\u200bFor comprehensive preparation in Calculus<\/strong> for the GATE 1996 Engineering Mathematics section, it’s beneficial to review previous years’ questions and their solutions.<\/span> Here are some resources that can aid your study:\u200b<\/p>\n

    \n
  1. \n

    ExamSIDE<\/strong>: This platform offers a collection of GATE Mechanical Engineering (ME) previous year questions on Calculus, organized subject-wise and chapter-wise, complete with detailed solutions.<\/span> \u200bGeeksforGeeks<\/span>+2<\/span><\/span>ExamSIDE<\/span>+2<\/span><\/span>ExamSIDE<\/span>+2<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

  2. \n

    GeeksforGeeks<\/strong>: They provide a compilation of GATE Computer Science and Engineering (CSE) previous year questions on Calculus, which can be instrumental in understanding the types of questions asked and the methodologies to solve them.<\/span> \u200bGeeksforGeeks<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

  3. \n

    GATE Overflow<\/strong>: This resource contains a comprehensive collection of previous year questions and solutions for GATE CSE, including those on Calculus. It serves as a valuable tool for in-depth understanding and practice.<\/span> \u200bGate Overflow<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ol>\n

    By thoroughly practicing these previous year questions and understanding the solutions, you can enhance your proficiency in Calculus and improve your performance in the GATE examination.<\/span><\/p>\n

    CSEIT – GATE 1996\/ Subject – Engineering Mathematics\/ Topic – Calulus The formula used to compu.<\/a><\/h3>\n
    \n
    \n
    \n

    Mathematics and engineering in computer science<\/a><\/h3>\n<\/div>\n<\/div>\n<\/div>\n

    SK Mondal’s – GATE Mathematics<\/a><\/h3>\n

    Here\u2019s a concise and exam-focused explanation of the GATE 1996 CSE\/IT \u2013 Engineering Mathematics<\/strong> topic related to Calculus<\/strong>, particularly focusing on the formula used to compute limits, derivatives, and integrals<\/strong> \u2014 often asked in various forms in GATE.<\/p>\n

    \n

    \ud83d\udcd8 Topic: Calculus \u2013 GATE 1996 CSE\/IT<\/strong><\/h2>\n
    \n

    \u2705 1. Limits \u2013 Formulae Used<\/strong><\/h3>\n

    Limits are used to understand the behavior of a function as it approaches a specific point.<\/p>\n

    \ud83d\udccc Standard Limit Results:<\/h4>\n

    lim\u2061x\u21920sin\u2061xx=1\\lim_{x \\to 0} \\frac{\\sin x}{x} = 1<\/span>x<\/span>\u2192<\/span>0<\/span><\/span>lim<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>x<\/span>sin<\/span>x<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>1<\/span><\/span><\/span><\/span><\/span> lim\u2061x\u219201\u2212cos\u2061xx2=12\\lim_{x \\to 0} \\frac{1 – \\cos x}{x^2} = \\frac{1}{2}<\/span>x<\/span>\u2192<\/span>0<\/span><\/span>lim<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>x<\/span>2<\/span><\/span><\/span><\/span>1\u2212<\/span>cos<\/span>x<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>21<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> lim\u2061x\u21920ex\u22121x=1\\lim_{x \\to 0} \\frac{e^x – 1}{x} = 1<\/span>x<\/span>\u2192<\/span>0<\/span><\/span>lim<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>x<\/span>e<\/span>x<\/span><\/span><\/span><\/span>\u2212<\/span>1<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>1<\/span><\/span><\/span><\/span><\/span> lim\u2061x\u2192\u221e(1+1x)x=e\\lim_{x \\to \\infty} \\left(1 + \\frac{1}{x}\\right)^x = e<\/span>x<\/span>\u2192<\/span>\u221e<\/span><\/span>lim<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>(<\/span><\/span>1<\/span>+<\/span>x<\/span>1<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>)<\/span><\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>e<\/span><\/span><\/span><\/span><\/span><\/p>\n

    \n

    \u2705 2. Derivatives \u2013 Important Rules<\/strong><\/h3>\n

    \ud83d\udd39 Definition of Derivative:<\/strong><\/h4>\n

    f\u2032(x)=lim\u2061h\u21920f(x+h)\u2212f(x)hf'(x) = \\lim_{h \\to 0} \\frac{f(x+h) – f(x)}{h}<\/span>f<\/span>\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span>(<\/span>x<\/span>)<\/span>=<\/span><\/span>h<\/span>\u2192<\/span>0<\/span><\/span>lim<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>h<\/span>f<\/span>(<\/span>x<\/span>+<\/span>h<\/span>)<\/span>\u2212<\/span>f<\/span>(<\/span>x<\/span>)<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \ud83d\udd39 Standard Derivatives:<\/strong><\/h4>\n

    ddx(xn)=nxn\u22121ddx(sin\u2061x)=cos\u2061xddx(cos\u2061x)=\u2212sin\u2061xddx(ex)=exddx(ln\u2061x)=1x\\frac{d}{dx} (x^n) = nx^{n-1} \\quad \\frac{d}{dx} (\\sin x) = \\cos x \\quad \\frac{d}{dx} (\\cos x) = -\\sin x \\quad \\frac{d}{dx} (e^x) = e^x \\quad \\frac{d}{dx} (\\ln x) = \\frac{1}{x}<\/span>d<\/span>x<\/span>d<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>(<\/span>x<\/span>n<\/span><\/span><\/span><\/span><\/span><\/span><\/span>)<\/span>=<\/span><\/span>n<\/span>x<\/span>n<\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span><\/span><\/span><\/span>d<\/span>x<\/span>d<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>(<\/span>sin<\/span>x<\/span>)<\/span>=<\/span><\/span>cos<\/span>x<\/span>d<\/span>x<\/span>d<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>(<\/span>cos<\/span>x<\/span>)<\/span>=<\/span><\/span>\u2212<\/span>sin<\/span>x<\/span>d<\/span>x<\/span>d<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>(<\/span>e<\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span>)<\/span>=<\/span><\/span>e<\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span>d<\/span>x<\/span>d<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>(<\/span>ln<\/span>x<\/span>)<\/span>=<\/span><\/span>x<\/span>1<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \ud83d\udd39 Product Rule:<\/strong><\/h4>\n

    ddx[u\u22c5v]=u\u2032v+uv\u2032\\frac{d}{dx}[u \\cdot v] = u’v + uv’<\/span>d<\/span>x<\/span>d<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>[<\/span>u<\/span>\u22c5<\/span><\/span>v<\/span>]<\/span>=<\/span><\/span>u<\/span>\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span>v<\/span>+<\/span><\/span>u<\/span>v<\/span>\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \ud83d\udd39 Quotient Rule:<\/strong><\/h4>\n

    ddx(uv)=u\u2032v\u2212uv\u2032v2\\frac{d}{dx}\\left(\\frac{u}{v}\\right) = \\frac{u’v – uv’}{v^2}<\/span>d<\/span>x<\/span>d<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>(<\/span><\/span>v<\/span>u<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span>=<\/span><\/span>v<\/span>2<\/span><\/span><\/span><\/span>u<\/span>\u2032<\/span><\/span><\/span><\/span>v<\/span>\u2212<\/span>u<\/span>v<\/span>\u2032<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \ud83d\udd39 Chain Rule:<\/strong><\/h4>\n

    ddx[f(g(x))]=f\u2032(g(x))\u22c5g\u2032(x)\\frac{d}{dx}[f(g(x))] = f'(g(x)) \\cdot g'(x)<\/span>d<\/span>x<\/span>d<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>[<\/span>f<\/span>(<\/span>g<\/span>(<\/span>x<\/span>))]<\/span>=<\/span><\/span>f<\/span>\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span>(<\/span>g<\/span>(<\/span>x<\/span>))<\/span>\u22c5<\/span><\/span>g<\/span>\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span>(<\/span>x<\/span>)<\/span><\/span><\/span><\/span><\/span><\/p>\n

    \n

    \u2705 3. Integration \u2013 Standard Formulae<\/strong><\/h3>\n

    \ud83d\udd39 Basic Formulas:<\/strong><\/h4>\n

    \u222bxndx=xn+1n+1+C(n\u2260\u22121)\\int x^n dx = \\frac{x^{n+1}}{n+1} + C \\quad (n \\ne -1)<\/span>\u222b<\/span>x<\/span>n<\/span><\/span><\/span><\/span><\/span><\/span><\/span>d<\/span>x<\/span>=<\/span><\/span>n<\/span>+<\/span>1x<\/span>n<\/span>+<\/span>1<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>C<\/span>(<\/span>n<\/span>\ue020<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>\u2212<\/span>1<\/span>)<\/span><\/span><\/span><\/span><\/span> \u222b1xdx=ln\u2061\u2223x\u2223+C\u222bexdx=ex+C\\int \\frac{1}{x} dx = \\ln |x| + C \\quad \\int e^x dx = e^x + C<\/span>\u222b<\/span>x<\/span>1<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>d<\/span>x<\/span>=<\/span><\/span>ln<\/span>\u2223<\/span>x<\/span>\u2223<\/span>+<\/span><\/span>C<\/span>\u222b<\/span>e<\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span>d<\/span>x<\/span>=<\/span><\/span>e<\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>C<\/span><\/span><\/span><\/span><\/span> \u222bsin\u2061xdx=\u2212cos\u2061x+C\u222bcos\u2061xdx=sin\u2061x+C\\int \\sin x dx = -\\cos x + C \\quad \\int \\cos x dx = \\sin x + C<\/span>\u222b<\/span>sin<\/span>x<\/span>d<\/span>x<\/span>=<\/span><\/span>\u2212<\/span>cos<\/span>x<\/span>+<\/span><\/span>C<\/span>\u222b<\/span>cos<\/span>x<\/span>d<\/span>x<\/span>=<\/span><\/span>sin<\/span>x<\/span>+<\/span><\/span>C<\/span><\/span><\/span><\/span><\/span><\/p>\n

    \ud83d\udd39 Integration by Parts:<\/strong><\/h4>\n

    \u222bu\u22c5v\u2009dx=u\u222bv\u2009dx\u2212\u222b(dudx\u22c5\u222bv\u2009dx)dx\\int u \\cdot v \\, dx = u \\int v \\, dx – \\int \\left( \\frac{du}{dx} \\cdot \\int v \\, dx \\right) dx<\/span>\u222b<\/span>u<\/span>\u22c5<\/span><\/span>v<\/span>d<\/span>x<\/span>=<\/span><\/span>u<\/span>\u222b<\/span>v<\/span>d<\/span>x<\/span>\u2212<\/span><\/span>\u222b<\/span>(<\/span><\/span>d<\/span>x<\/span>d<\/span>u<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u22c5<\/span>\u222b<\/span>v<\/span>d<\/span>x<\/span>)<\/span><\/span><\/span>d<\/span>x<\/span><\/span><\/span><\/span><\/span><\/p>\n

    \ud83d\udd39 Definite Integration:<\/strong><\/h4>\n

    \u222babf(x)dx=F(b)\u2212F(a)(where\u00a0F(x)\u00a0is\u00a0antiderivative\u00a0of\u00a0f(x))\\int_a^b f(x) dx = F(b) – F(a) \\quad \\text{(where \\( F(x) \\) is antiderivative of \\( f(x) \\))}<\/span>\u222b<\/span>a<\/span><\/span>b<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>f<\/span>(<\/span>x<\/span>)<\/span>d<\/span>x<\/span>=<\/span><\/span>F<\/span>(<\/span>b<\/span>)<\/span>\u2212<\/span><\/span>F<\/span>(<\/span>a<\/span>)<\/span>(where\u00a0<\/span>F<\/span>(<\/span>x<\/span>)<\/span>\u00a0is\u00a0antiderivative\u00a0of\u00a0<\/span>f<\/span>(<\/span>x<\/span>)<\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \n

    \ud83c\udfaf GATE 1996 Example Question (Conceptual)<\/strong><\/h2>\n

    Q:<\/strong>
    Evaluate:<\/p>\n

    lim\u2061x\u21920ex\u2212sin\u2061x\u2212xx3\\lim_{x \\to 0} \\frac{e^x – \\sin x – x}{x^3}<\/span>x<\/span>\u2192<\/span>0<\/span><\/span>lim<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>x<\/span>3<\/span><\/span><\/span><\/span>e<\/span>x<\/span><\/span><\/span><\/span>\u2212<\/span>sin<\/span>x<\/span>\u2212<\/span>x<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \u2705 Solution Outline:<\/h3>\n

    Use Taylor Series Expansion<\/strong>:<\/p>\n

    ex=1+x+x22!+x33!+\u22efe^x = 1 + x + \\frac{x^2}{2!} + \\frac{x^3}{3!} + \\cdots <\/span>e<\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>1<\/span>+<\/span><\/span>x<\/span>+<\/span><\/span>2!<\/span>x<\/span>2<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>3!<\/span>x<\/span>3<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>\u22ef<\/span><\/span><\/span><\/span><\/span> sin\u2061x=x\u2212x33!+\u22ef\\sin x = x – \\frac{x^3}{3!} + \\cdots<\/span>sin<\/span>x<\/span>=<\/span><\/span>x<\/span>\u2212<\/span><\/span>3!<\/span>x<\/span>3<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>\u22ef<\/span><\/span><\/span><\/span><\/span><\/p>\n

    So:<\/p>\n

    ex\u2212sin\u2061x\u2212x=(1+x+x22+x36+\u2026\u2009)\u2212(x\u2212x36+\u2026\u2009)\u2212x=1+x22+x33e^x – \\sin x – x = (1 + x + \\frac{x^2}{2} + \\frac{x^3}{6} + \\dots) – (x – \\frac{x^3}{6} + \\dots) – x = 1 + \\frac{x^2}{2} + \\frac{x^3}{3}<\/span>e<\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>sin<\/span>x<\/span>\u2212<\/span><\/span>x<\/span>=<\/span><\/span>(<\/span>1<\/span>+<\/span><\/span>x<\/span>+<\/span><\/span>2x<\/span>2<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>6x<\/span>3<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>\u2026<\/span>)<\/span>\u2212<\/span><\/span>(<\/span>x<\/span>\u2212<\/span><\/span>6x<\/span>3<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>\u2026<\/span>)<\/span>\u2212<\/span><\/span>x<\/span>=<\/span><\/span>1<\/span>+<\/span><\/span>2x<\/span>2<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>3x<\/span>3<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    Now divide by x3x^3<\/span>x<\/span>3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>:<\/p>\n

    1+x22+x33x3=1×3+12x+13\u2192\u221e\\frac{1 + \\frac{x^2}{2} + \\frac{x^3}{3}}{x^3} = \\frac{1}{x^3} + \\frac{1}{2x} + \\frac{1}{3} \\to \\infty<\/span>x<\/span>3<\/span><\/span><\/span><\/span>1+<\/span>2<\/span><\/span>x<\/span>2<\/span><\/span><\/span><\/span><\/span>\u200b<\/span>+<\/span>3<\/span><\/span>x<\/span>3<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>x<\/span>3<\/span><\/span><\/span><\/span>1<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>2x<\/span>1<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>31<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u2192<\/span><\/span>\u221e<\/span><\/span><\/span><\/span><\/span><\/p>\n

    But since numerator starts with constant term, the limit does not exist as finite<\/strong>.<\/p>\n

    (However, if the problem had been lim\u2061x\u21920ex\u2212sin\u2061x\u2212xx2\\lim_{x \\to 0} \\frac{e^x – \\sin x – x}{x^2}<\/span>limx<\/span>\u2192<\/span>0<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>x<\/span>2<\/span><\/span><\/span><\/span><\/span>e<\/span>x<\/span><\/span><\/span><\/span>\u2212<\/span>s<\/span>i<\/span>n<\/span><\/span>x<\/span>\u2212<\/span>x<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, then the approach would give a finite value.)<\/p>\n

    \n

    \u2705 Summary Table<\/h2>\n
    \n
    \n\n\n\n\n\n\n\n
    Concept<\/th>\nFormula Used<\/th>\n<\/tr>\n<\/thead>\n
    Limit<\/td>\nsin\u2061xx=1\\frac{\\sin x}{x} = 1<\/span>x<\/span><\/span><\/span>s<\/span>i<\/span>n<\/span><\/span>x<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>1<\/span><\/span><\/span><\/span>, Series forms<\/td>\n<\/tr>\n
    Derivative<\/td>\nChain Rule, Product Rule, Standard forms<\/td>\n<\/tr>\n
    Integration<\/td>\nPower Rule, By Parts, Definite Integrals<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
    \n
    <\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n

    Would you like:<\/p>\n