# 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