Here, you will learn what is squeeze theorem or sandwich theorem of limit and limit of exponential function with examples.
Letโs begin โ
Squeeze Theorem (Sandwich Theorem)
If f(x) \(\leq\) g(x) \(\leq\) h(x); \(\forall\) x in the neighbourhood at x = a and
\(\displaystyle{\lim_{x \to a}}\) f(x) = l = \(\displaystyle{\lim_{x \to 1}}\) h(x) then \(\displaystyle{\lim_{x \to 1}}\) g(x) = l
Example : Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^2sin {1\over x}\) = 0
Solution :
\(sin ({1\over x})\) lies between -1 and 1.
\(\implies\) \(-x^2\) \(\leq\) \(x^2sin {1\over x}\) \(\leq\) \(x^2\)
\(\implies\) \(\displaystyle{\lim_{x \to 0}}\) \(x^2sin {1\over x}\) = 0 as
\(\displaystyle{\lim_{x \to 0}}\) \(-x^2\) = \(\displaystyle{\lim_{x \to 0}}\) \(x^2\) = 0
Limit of Exponential Functions
(a)ย \(\displaystyle{\lim_{x \to 0}}\) \(a^x โ 1\over x\) = lna (a > 0) In particular \(\displaystyle{\lim_{x \to 0}}\) \(e^x โ 1\over x\) = 1
In general if \(\displaystyle{\lim_{x \to a}}\) f(x) = 0, then \(\displaystyle{\lim_{x \to a}}\) \(a^{f(x) โ 1}\over {f(x)}\) = lna, a > 0
(b)ย (i)ย \(\displaystyle{\lim_{x \to 0}}\) \({(1 + x)}^{1\over x}\) = e = \(\displaystyle{\lim_{x \to \infty}}\) \({(1 + {1\over x})}^x\) (The base and exponent depends on the same variable.)
In general if \(\displaystyle{\lim_{x \to a}}\) f(x) = 0, then \(\displaystyle{\lim_{x \to a}}\) \({(1 + f(x))}^{1/f(x)}\) = e
(ii)ย \(\displaystyle{\lim_{x \to 0}}\) \(ln(1 + x)\over x\) = 1
(iii)ย 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) โ 1]
(c)ย If \(\displaystyle{\lim_{x \to a}}\) f(x) = A > 0 & \(\displaystyle{\lim_{x \to a}}\) \(\phi(x)\) = B, then \(\displaystyle{\lim_{x \to a}}\) \({[f(x)]}^{\phi(x)}\) = \(e^{B lnA}\) = \(A^B\)
Example : Evaluate : \(\displaystyle{\lim_{x \to 1}}\) \({(log_3 3x)}^{log_x 3}\)
Solution :
\(\displaystyle{\lim_{x \to 1}}\) \({(log_3 3x)}^{log_x 3}\) = \(\displaystyle{\lim_{x \to 1}}\)
\({(log_3 3 + log_3 x)}^{log_x 3}\)
= \(\displaystyle{\lim_{x \to 1}}\) \({(1 + log_3 x)}^{1/log_3 x}\) = e
