Решения NCERT для класса 11. Глава 13. Пределы и производные. Упражнение 13.1 | Набор 2

Вопрос 17:

Решение:

In, as x⇢0

As we know, cos 2θ = 1-2sin²θ

Substituting the values, we get

=

Put x = 0, we get

As, this limit becomes undefined

Now, let’s multiply and divide the numerator by x² and denominator byto make it equivalent to theorem.

Hence, we have

=

=

By using the theorem, we get

=

=

=

= 4

Вопрос 18:

Решение:

In, as x⇢0

Put x = 0, we get

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem.

Hence, we have

=

By using the theorem, we get

=

=

Putting x=0, we have

=

Вопрос 19:

Решение:

In, as x⇢0

Put x = 0, we get

= 0 ×1

= 0

Вопрос 20:

Решение:

In, as x⇢0

Put x = 0, we get

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem.

Hence, we can write the equation as follows:

=

By using the theorem, we get

=

=

=

=

Putting x=0, we have

= 1

Вопрос 21:

Решение:

In, as x⇢0

By simplification, we get

Put x = 0, we get

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem:

By using the trigonometric identities,

cos 2θ = 1-2sin²θ

sin 2θ = 2 sinθ cosθ

Hence, we can write the equation as follows:

=

=

Putting x=0, we have

= 0

Вопрос 22:

Решение:

In, as x⇢

Put x =, we get

As, this limit becomes undefined

Now, let’s simplify the equation :

Let’s take

As, x⇢⇒ p⇢0

Hence, we can write the equation as follows:

=

=(As tan (π+θ) = tan θ)

=

=

Now, let’s multiply and divide the equation by 2 to make it equivalent to theorem

=

=

As p⇢0, then 2p⇢0

=

Using the theorem and putting p=0, we have

= 2×1×1

= 2

Вопрос 23: Найдите а также , куда

Решение:

Let’s calculate, the limits when x⇢0

Here,

Left limit =

Right limit =

Limit value =

Hence,, then limit exists

Now, let’s calculate, the limits when x⇢1

Here,

Left limit =

Right limit =

Limit value =

Hence,, then limit exists

Вопрос 24: Найдите , куда

Решение:

Let’s calculate, the limits when x⇢1

Here,

Left limit =

Right limit =

As,

Hence, limit does not exists when x⇢1.

Вопрос 25: Оцените , куда

Решение:

Let’s calculate, the limits when x⇢0

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =

Right limit =

As,

Hence, limit does not exists when x⇢0.

Вопрос 26: Найдите , куда

Решение:

Let’s calculate, the limits when x⇢0

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =

Right limit =

As,

Hence, limit does not exists when x⇢0.

Вопрос 27: Найдите , где f(x)=|x|-5.

Решение:

Let’s calculate, the limits when x⇢5

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =

Right limit =

Hence,, then limit exists

Вопрос 28: Предположим и если каковы возможные значения a и b?

Решение:

As, it is given

Let’s calculate, the limits when x⇢1

Here,

Left limit =

Right limit =

Limit value f(1) = 4

So, as limit exists then it should satisfy

Hence, a+b = 4 and b-a = 4

Solving these equation, we get

a = 0 and b = 4

Вопрос 29: Пусть a ₁ , a ₂ , . . ., a _n — фиксированные действительные числа и определяют функцию

f(x) = (xa ₁ ) (xa ₂ )………… (x’s).

Что такое ? Для некоторого a ≠ a ₁ , a ₂ , …, an вычислить .

Решение:

Here, f(x) = (x-a₁) (x-a₂)………… (x-a_n).

Then,

=

= (a₁-a₁) (a₁-a₂)………… (a₁-a_n)

= 0

Now, let’s calculate for

=

= (a-a₁) (a-a₂)………… (a-a_n)

= (a-a₁) (a-a₂)………… (a-a_n)

Вопрос 30: Если

Для какого значения a делает существуют?

Решение:

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Let’s check for three cases of a:

• When a=0

Let’s calculate, the limits when x⇢0

Left limit =

Right limit =

As,

Hence, limit does not exists when x⇢0.

• When a>0

Let’s take a=2, for reference

Let’s calculate, the limits when x⇢2

Left limit =

Right limit =

As,

Hence, limit exists when x⇢2.

• When a<0

Let’s take a=-2, for reference

Let’s calculate, the limits when x⇢ -2

Left limit =

Right limit =

As,

Hence, limit exists when x⇢ -2.

Вопрос 31: Если функция f(x) удовлетворяет условию , оценивать

Решение:

Here, as it is given

Put x = 1 in RHS, we get

= 2

Hence proved!

Вопрос 32: Если . Для каких целых чисел m и n оба а также существуют?

Решение:

Let’s calculate, the limits when x⇢0

Here,

Left limit =

Right limit =

Hence,

, then limit exists

m = n

Now, let’s calculate, the limits when x⇢1

Here,

Left limit =

Right limit =

Hence,, then limit exists.