Informative line

Theorems

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

Intermediate Value Theorem

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 \(\alpha\).

Illustration Questions

Consider \(f(x) =2x^3+x^2+4x–6\) in \([1,\,2]\) Which of the following values must be taken by 'f' when \(x\) lies in the interval \([1,\,2]\). 

A –81

B –7

C 10

D 78

×

\(f(1)=2+1+4–6=1\)

\(f(2)=2×8+4+8–6=22\)

\(f\) is continuous because of being polynomial.

\(f\) must take all values in \([1,\,22]\,\to\) Intermediate value theorem

10 is the Answer.

The other values may or may not be taken by the function.

Consider \(f(x) =2x^3+x^2+4x–6\) in \([1,\,2]\) Which of the following values must be taken by 'f' when \(x\) lies in the interval \([1,\,2]\). 

A

–81

.

B

–7

C

10

D

78

Option C is Correct

Application of Intermediate Value Theorem

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.

Illustration Questions

Consider the function \(f(x)=\sqrt x–x\) then \(f(x)=0\) must have at least one root in 

A [4, 9]

B [9, 16]

C [16, 25]

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

×

For option a 

\(f(4)=\sqrt 4-4=2-4=-2\)

\(f(9)=\sqrt 9-9=3-9=-6\)

\(f(4),\,f(9)\) are of same sign.

For option b

\(f(9)=\sqrt 9-9=3-9=-6\)

\(f(16)=\sqrt 16 -16=-12\)

\(f(9),\,f(16)\) are of same sign

For option c

\(f(16)=\sqrt 16 -16=-12\)

\(f(25)=5-25=-20\)

\(f(16),\,f(25)\) are of same sign

For option d

\(f\left(\dfrac{1}{2}\right)=\dfrac{1}{\sqrt 2}-\dfrac{1}{2}=0.71-0.5=0.21\)

\(f(4)=\sqrt 4-4=2-4=-2\)

\(f\left(\dfrac{1}{2}\right),\,f(4)\) are of opposite sign 

\(\therefore\,f(x)=0\) must have a root in \(\left(\dfrac{1}{2},\,4\right)\)

Consider the function \(f(x)=\sqrt x–x\) then \(f(x)=0\) must have at least one root in 

A

[4, 9]

.

B

[9, 16]

C

[16, 25]

D

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

Option D is Correct

Theorem

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.

Illustration Questions

Let \(f(x) = 2x^2 + 5x –7\) and g(x) be a function such that g(x) > f(x) for all real x, then \(\lim\limits_{x\rightarrow2}\;g(x)\) can take which of the following values ?

A 30

B 2

C –11

D –4

×

By the theorem,

\(g(x)>f(x)\Rightarrow \lim\limits_{x\rightarrow 2}\;g(x)>\lim\limits_{x\rightarrow 2}\;f(x)\)

\(\Rightarrow \lim\limits_{x\rightarrow 2}\;g(x)>\lim\limits_{x\rightarrow 2}\;(2x^2+5x-7)\)

\(\Rightarrow \lim\limits_{x\rightarrow 2}\;g(x)>2×4+5×2–7\)

\(\Rightarrow \lim\limits_{x\rightarrow 2}\;g(x)>11\) ( \(\lim\limits_{x\rightarrow 2}\;g(x)\) can be any number greater than 11)

So, correct option is 'a' .                (look for a number greater than 11)

Let \(f(x) = 2x^2 + 5x –7\) and g(x) be a function such that g(x) > f(x) for all real x, then \(\lim\limits_{x\rightarrow2}\;g(x)\) can take which of the following values ?

A

30

.

B

2

C

–11

D

–4

Option A is Correct

Squeeze Theorem or Sandwich Theorem

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.

Illustration Questions

The value of the limit  \(\lim\limits_{x\rightarrow0}\;x^3\,sin\dfrac {2}{x}\) is

A 2

B 0

C 17

D –8

×

Since \(sin\dfrac {2}{x}\) lies between –1 and 1.

\(\Rightarrow\) \(-1\leq sin\dfrac {2}{x}\leq 1\)

\(\Rightarrow -x^3 \leq x^3sin\dfrac {2}{x} \leq x^3\)

We thought of two function one of which is greater than the given function and other is less, and both have the same limit at x = 0

Now, 

\(\lim\limits_{x\rightarrow0}\;x^3\,=0\) and \(\lim\limits_{x\rightarrow0}\;-x^3=0\), so by squeeze theorem the given limit is also 0.

The value of the limit  \(\lim\limits_{x\rightarrow0}\;x^3\,sin\dfrac {2}{x}\) is

A

2

.

B

0

C

17

D

–8

Option B is Correct

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