What is Squeeze Theorem – Limit of Exponential Functions

Here, you will learn what is squeeze theorem or sandwich theorem of limit and limit of exponential function with examples.

Let’s begin –

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)  \(\displaystyle{\lim_{x \to 0}}\) \(a^x – 1\over x\) = lna (a > 0) In particular \(\displaystyle{\lim_{x \to 0}}\) \(e^x – 1\over x\) = 1

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

(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.)

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)  \(\displaystyle{\lim_{x \to 0}}\) \(ln(1 + x)\over x\) = 1

(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]

(c)  If \(\displaystyle{\lim_{x \to a}}\) f(x) = A > 0 & \(\displaystyle{\lim_{x \to a}}\) \(\phi(x)\) = B, then \(\displaystyle{\lim_{x \to a}}\) \({[f(x)]}^{\phi(x)}\) = \(e^{B lnA}\) = \(A^B\)

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

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