Learn sandwich or squeeze theorem calculus, practice squeeze theorem limits problems and intermediate value theorem calculus.

Let \('f'\) be a continuous function in \([a,\,b]\) and N be any number between \(f(a)\) and \(f(b)\), then there exists at least one \(\alpha \in(a,\,b)\) such that. \(f(\alpha)=N\)

- There is no restriction on the number of values of \(\alpha\).

If \('f'\) is a continuous function in [a, b] and \(f(a)\,f(b)<0\)

i.e \(f(a)\) and \(f(b)\) are of opposite signs then there is at least one root of \(f(x)=0\) in [a, b]

- Graph must intersect the \(x\) axis at least once.

A [4, 9]

B [9, 16]

C [16, 25]

D \(\left[\dfrac{1}{2},\,4\right]\)

If \(f(x)\leq g(x)\) for values of x which are near 'a' then

\(\lim\limits_{x\rightarrow a}f(x)\leq\lim\limits_{x\rightarrow a}g(x)\) provided both these limit exist.

This means that if f takes values larger than g for all value of x then the limit of larger function i.e. f is also greater than limit of g.

Let there be 3 functions, 'f' , 'g' and 'h' such that

\(f(x)\leq g(x) \leq h(x)\) for values of x near 'a' and

\(\lim\limits_{x\rightarrow a}\;f(x)=\lim\limits_{x\rightarrow a}\;h(x)=\ell\) , then \(\lim\limits_{x\rightarrow a}\;g(x)=\ell\)

- We say that the value of g(x) gets sandwiched between those of f(x) and h(x), therefore the name of theorem.
- Look at the graph.