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 $$\alpha$$.

#### 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.

#### 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.

#### 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.

#### 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