QuestionsQuestion 1Find the limit of the following sequences if exists:(i)an=2n+3n+1a_n = \dfrac{2n+3}{n+1}an=n+12n+3(ii)bn=2n+3n2+1b_n = \dfrac{2n+3}{n^2+1}bn=n2+12n+3(iii)cn=5n22n+3c_n = \dfrac{5n^2}{2n+3}cn=2n+35n2(iv)dn=n2−3n+12n2+n+4d_n = \dfrac{n^2 - 3n + 1}{2n^2 + n + 4}dn=2n2+n+4n2−3n+1SolutionTheoryQuestion 2Evaluate each limit by using theorems of limits:(i)limx→3(2x+4)\lim_{x \to 3}(2x + 4)limx→3(2x+4)(ii)limx→1(3x2−2x+4)\lim_{x \to 1}(3x^2 - 2x + 4)limx→1(3x2−2x+4)(iii)limx→5x2+x+4\lim_{x \to 5}\sqrt{x^2 + x + 4}limx→5x2+x+4(iv)limx→2x2+4\lim_{x \to 2}\sqrt{x^2 + 4}limx→2x2+4(v)limx→2(x3+1−x2+5)\lim_{x \to 2}(\sqrt{x^3 + 1} - \sqrt{x^2 + 5})limx→2(x3+1−x2+5)(vi)limx→22x3+5x3x−2\lim_{x \to 2}\dfrac{2x^3 + 5x}{3x - 2}limx→23x−22x3+5xSolutionTheoryQuestion 3Evaluate each limit by using algebraic techniques:(i)limx→1x3−xx+1\lim_{x \to 1}\dfrac{x^3 - x}{x + 1}limx→1x+1x3−x(ii)limx→3(x2−5x+6x2−2x−3)\lim_{x \to 3}\left(\dfrac{x^2 - 5x + 6}{x^2 - 2x - 3}\right)limx→3(x2−2x−3x2−5x+6)(iii)limx→2(x3−8x2−5x+6)\lim_{x \to 2}\left(\dfrac{x^3 - 8}{x^2 - 5x + 6}\right)limx→2(x2−5x+6x3−8)(iv)limx→1x3−3x2+3x−1x3−x\lim_{x \to 1}\dfrac{x^3 - 3x^2 + 3x - 1}{x^3 - x}limx→1x3−xx3−3x2+3x−1(v)limx→2(x3−6x2+12x−8x3−4x)\lim_{x \to 2}\left(\dfrac{x^3 - 6x^2 + 12x - 8}{x^3 - 4x}\right)limx→2(x3−4xx3−6x2+12x−8)(vi)limx→1(x4−1x2−3x+2)\lim_{x \to 1}\left(\dfrac{x^4 - 1}{x^2 - 3x + 2}\right)limx→1(x2−3x+2x4−1)(vii)limx→2x−2x+2−6−x\lim_{x \to 2}\dfrac{x - 2}{\sqrt{x + 2} - \sqrt{6 - x}}limx→2x+2−6−xx−2(viii)limh→0x+h−xh\lim_{h \to 0}\dfrac{\sqrt{x + h} - \sqrt{x}}{h}limh→0hx+h−x(ix)limx→axn−anxm−am\lim_{x \to a}\dfrac{x^n - a^n}{x^m - a^m}limx→axm−amxn−anSolutionTheoryQuestion 4Evaluate the following limits:(i)limx→0sin5xx\lim_{x \to 0}\dfrac{\sin 5x}{x}limx→0xsin5x(ii)limx→0sinx2x\lim_{x \to 0}\dfrac{\sin x^2}{x}limx→0xsinx2(iii)limθ→01−cosθsinθ\lim_{\theta \to 0}\dfrac{1 - \cos\theta}{\sin\theta}limθ→0sinθ1−cosθ(iv)limx→π4sinx−cosxx−π4\lim_{x \to \frac{\pi}{4}}\dfrac{\sin x - \cos x}{x - \frac{\pi}{4}}limx→4πx−4πsinx−cosx(v)limx→0cosax−cosbxx2\lim_{x \to 0}\dfrac{\cos ax - \cos bx}{x^2}limx→0x2cosax−cosbx(vi)limx→π4tanx−1x−π4\lim_{x \to \frac{\pi}{4}}\dfrac{\tan x - 1}{x - \frac{\pi}{4}}limx→4πx−4πtanx−1(vii)limx→01−cos2xx2\lim_{x \to 0}\dfrac{1 - \cos 2x}{x^2}limx→0x21−cos2x(viii)limx→0cosax−cosbxcoscx−cosdx\lim_{x \to 0}\dfrac{\cos ax - \cos bx}{\cos cx - \cos dx}limx→0coscx−cosdxcosax−cosbx(ix)limx→1x3−1x2−1\lim_{x \to 1}\dfrac{x^3 - 1}{x^2 - 1}limx→1x2−1x3−1(x)limx→3x2−xlogx+3logx−9x−3\lim_{x \to 3}\dfrac{x^2 - x\log x + 3\log x - 9}{x - 3}limx→3x−3x2−xlogx+3logx−9(xi)limx→0x(2x−1)1−cosx\lim_{x \to 0}\dfrac{x(2^x - 1)}{1 - \cos x}limx→01−cosxx(2x−1)SolutionTheoryQuestion 5Express each limit in terms of eee.(i)limn→∞(1+1n)2n\lim_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^{2n}limn→∞(1+n1)2n(ii)limn→+∞(1+1n)n2\lim_{n \to +\infty}\left(1 + \dfrac{1}{n}\right)^{\frac{n}{2}}limn→+∞(1+n1)2n(iii)limn→+∞(1−1n)n\lim_{n \to +\infty}\left(1 - \dfrac{1}{n}\right)^nlimn→+∞(1−n1)n(iv)limn→+∞(1+13n)n\lim_{n \to +\infty}\left(1 + \dfrac{1}{3n}\right)^nlimn→+∞(1+3n1)n(v)limn→+∞(1+4n)n\lim_{n \to +\infty}\left(1 + \dfrac{4}{n}\right)^nlimn→+∞(1+n4)n(vi)limx→0(1+3x)2x\lim_{x \to 0}(1 + 3x)^{\frac{2}{x}}limx→0(1+3x)x2(vii)limx→0(1+2x2)1x2\lim_{x \to 0}(1 + 2x^2)^{\frac{1}{x^2}}limx→0(1+2x2)x21(viii)limx→0eax−ebxabx\lim_{x \to 0}\dfrac{e^{ax} - e^{bx}}{abx}limx→0abxeax−ebx(ix)limx→∞(x1+x)x\lim_{x \to \infty}\left(\dfrac{x}{1 + x}\right)^xlimx→∞(1+xx)x(x)limx→0ex−1ex+1\lim_{x \to 0}\dfrac{e^x - 1}{e^x + 1}limx→0ex+1ex−1, x<0x < 0x<0(xi)limx→0ex−1ex+1\lim_{x \to 0}\dfrac{e^x - 1}{e^x + 1}limx→0ex+1ex−1, x>0x > 0x>0(xii)limx→2ex−e2x−2\lim_{x \to 2}\dfrac{e^x - e^2}{x - 2}limx→2x−2ex−e2SolutionTheory