Example 1 : <\/span>If \\(\\displaystyle{\\lim_{x \\to \\infty}}\\)(\\({x^3+1\\over x^2+1}-(ax+b)\\)) = 2, then find the value of a and b.<\/p>\n Solution : <\/span>\t\\(\\displaystyle{\\lim_{x \\to \\infty}}\\)(\\({x^3+1\\over x^2+1}-(ax+b)\\)) = 2 Example 2 : <\/span> Evaluate : \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\((2+x)sin(2+x)-2sin2\\over x\\)<\/p>\n Solution : <\/span>\\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(2(sin(2+x)-sin2)+xsin(2+x)\\over x\\) Example 3 : <\/span>Evaluate : \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(xln(1+2tanx)\\over 1-cosx\\)<\/p>\n Solution : <\/span>\\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(xln(1+2tanx)\\over 1-cosx\\) Example 4 : <\/span>Evaluate : \\(\\displaystyle{\\lim_{x \\to \\infty}}\\) \\(({7x^2+1\\over 5x^2-1})^{x^5\\over {1-x^3}}\\)<\/p>\n Solution : <\/span>Here f(x) = \\({7x^2+1\\over 5x^2-1}\\) Practice these given limits examples to test your knowledge on concepts of limits.<\/p> \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some limits examples for better understanding of limit concepts. Example 1 : If \\(\\displaystyle{\\lim_{x \\to \\infty}}\\)(\\({x^3+1\\over x^2+1}-(ax+b)\\)) = 2, then find the value of a and b. Solution : \\(\\displaystyle{\\lim_{x \\to \\infty}}\\)(\\({x^3+1\\over x^2+1}-(ax+b)\\)) = 2 \\(\\implies\\) \\(\\displaystyle{\\lim_{x \\to \\infty}}\\)\\(x^3(1-a)-bx^2-ax+(1-b)\\over x^2+1\\) = 2 \\(\\implies\\) \\(\\displaystyle{\\lim_{x \\to \\infty}}\\)\\(x(1-a)-b-{a\\over x}+{(1-b)\\over x^2}\\over 1+{1\\over x^2}\\) = …<\/p>\n
\\(\\implies\\) \\(\\displaystyle{\\lim_{x \\to \\infty}}\\)\\(x^3(1-a)-bx^2-ax+(1-b)\\over x^2+1\\) = 2
\\(\\implies\\) \\(\\displaystyle{\\lim_{x \\to \\infty}}\\)\\(x(1-a)-b-{a\\over x}+{(1-b)\\over x^2}\\over 1+{1\\over x^2}\\) = 2
\\(\\implies\\) 1 – a = 0, -b = 2 \\(\\implies\\) a = 1, b = -2<\/p>\n
\n\n \n
\n = \\(\\displaystyle{\\lim_{x \\to 0}}\\)(\\(2.2.cos(2+{x\\over 2})sin{x\\over 2}\\over x\\) + sin(2+x))
= \\(\\displaystyle{\\lim_{x \\to 0}}\\)\\(2cos(2+{x\\over 2})sin{x\\over 2}\\over {x\\over 2}\\) + \\(\\displaystyle{\\lim_{x \\to 0}}\\)sin(2+x)
= 2cos2 + sin2\n <\/p>
\n\n \n
= \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(xln(1+2tanx)\\over {1-cosx\\over x^2}.x^2\\).\\(2tanx\\over 2tanx\\)
= 4<\/p>\n
\n\n
\\(\\phi\\)(x) = \\({x^5\\over {1-x^3}}\\) = \\(x^2x^3\\over 1-x^3\\) = \\(x^2\\over {1\\over x^3}-1\\)
\\(\\therefore\\) \\(\\displaystyle{\\lim_{x \\to \\infty}}\\) f(x) = \\(7\\over 5\\) & \\(\\displaystyle{\\lim_{x \\to \\infty}}\\) \\(\\phi\\)(x) \\(\\rightarrow\\) – \\(\\infty\\)
\\(\\implies\\) \\(\\displaystyle{\\lim_{x \\to \\infty}}\\) \\((f(x))^{\\phi (x)}\\) = \\(({7\\over 5})^{-\\infty}\\) = 0<\/p>\n
\n\n