{"id":3961,"date":"2021-08-11T13:12:22","date_gmt":"2021-08-11T13:12:22","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3961"},"modified":"2021-11-26T16:35:56","modified_gmt":"2021-11-26T11:05:56","slug":"what-is-squeeze-theorem","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-squeeze-theorem\/","title":{"rendered":"What is Squeeze Theorem – Limit of Exponential Functions"},"content":{"rendered":"
Here, you will learn what is squeeze theorem or sandwich theorem of limit and limit of exponential function with examples.<\/p>\n
Let’s begin –<\/p>\n
\nIf f(x) \\(\\leq\\) g(x) \\(\\leq\\) h(x); \\(\\forall\\) x in the neighbourhood at x = a and<\/p>\n
\\(\\displaystyle{\\lim_{x \\to a}}\\) f(x) = l = \\(\\displaystyle{\\lim_{x \\to 1}}\\) h(x) then \\(\\displaystyle{\\lim_{x \\to 1}}\\) g(x) = l<\/p>\n<\/blockquote>\n\n\n
Example : <\/span>Evaluate : \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(x^2sin {1\\over x}\\) = 0 <\/p>\n
Solution : <\/span>\n \\(sin ({1\\over x})\\) lies between -1 and 1.
\n\t\t \\(\\implies\\) \\(-x^2\\) \\(\\leq\\) \\(x^2sin {1\\over x}\\) \\(\\leq\\) \\(x^2\\)
\n\t\t \\(\\implies\\) \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(x^2sin {1\\over x}\\) = 0 as \n\t\t \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(-x^2\\) = \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(x^2\\) = 0
<\/p>\n\n\nLimit of Exponential Functions<\/h2>\n
(a) \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(a^x – 1\\over x\\) = lna (a > 0) In particular \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(e^x – 1\\over x\\) = 1<\/p>\n
In general if \\(\\displaystyle{\\lim_{x \\to a}}\\) f(x) = 0, then \\(\\displaystyle{\\lim_{x \\to a}}\\) \\(a^{f(x) – 1}\\over {f(x)}\\) = lna, a > 0<\/p>\n
(b) (i) \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\({(1 + x)}^{1\\over x}\\) = e = \\(\\displaystyle{\\lim_{x \\to \\infty}}\\) \\({(1 + {1\\over x})}^x\\) (The base and exponent depends on the same variable.)<\/p>\n
In general if \\(\\displaystyle{\\lim_{x \\to a}}\\) f(x) = 0, then \\(\\displaystyle{\\lim_{x \\to a}}\\) \\({(1 + f(x))}^{1\/f(x)}\\) = e<\/p>\n
(ii) \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(ln(1 + x)\\over x\\) = 1<\/p>\n
(iii) If \\(\\displaystyle{\\lim_{x \\to a}}\\) f(x) = 1 and \\(\\displaystyle{\\lim_{x \\to a}}\\) \\(\\phi(x)\\) = \\(\\infty\\) then; \\(\\displaystyle{\\lim_{x \\to a}}\\) \\({[f(x)]}^{\\phi(x)}\\) = \\(e^k\\); where k = \\(\\displaystyle{\\lim_{x \\to a}}\\) \\(\\phi(x)\\) [f(x) – 1]<\/p>\n
(c) If \\(\\displaystyle{\\lim_{x \\to a}}\\) f(x) = A > 0 & \\(\\displaystyle{\\lim_{x \\to a}}\\) \\(\\phi(x)\\) = B, then \\(\\displaystyle{\\lim_{x \\to a}}\\) \\({[f(x)]}^{\\phi(x)}\\) = \\(e^{B lnA}\\) = \\(A^B\\)<\/p>\n\n\n
Example : <\/span>Evaluate : \\(\\displaystyle{\\lim_{x \\to 1}}\\) \\({(log_3 3x)}^{log_x 3}\\) <\/p>\n
Solution : <\/span>\n \\(\\displaystyle{\\lim_{x \\to 1}}\\) \\({(log_3 3x)}^{log_x 3}\\) = \\(\\displaystyle{\\lim_{x \\to 1}}\\)\n \\({(log_3 3 + log_3 x)}^{log_x 3}\\)
= \\(\\displaystyle{\\lim_{x \\to 1}}\\) \\({(1 + log_3 x)}^{1\/log_3 x}\\) = e
<\/p>\n\n\n\n