Evaluate the following limit : limx→0[1+x3-1+xx]

Evaluate the following limit :

`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`

योग

`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`

= `lim_(x -> 0) ((1 + x)^(1/3) - (1 + x)^(1/2))/x`

Put 1 + x = y ⇒ x = y − 1

As x→0, y→1

∴ `lim_(x -> 0) ((1 + x)^(1/3) - (1 + x)^(1/2))/x`

= `lim_(y -> 1) (y^(1/3) - y^(1/2))/(y - 1)`

= `lim_(y -> 1) ((y^(1/3) - 1) - (y^(1/2) - 1))/(y - 1)`

= `lim_(y -> 1) ((y^(1/3) - 1)/(y - 1) - y^(1/2 - 1)/(y - 1))`

= `lim_(y -> 1) (y^(1/3) - 1^(1/3))/(y - 1) - lim_(y -> 1) (y^(1/2) - 1^(1/2))/(y - 1)`

= `1/3(1)^(-(2)/3) - 1/2(1)^(-(1)/2) ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`

= `1/3 - 1/2`

= `(2 - 3)/6`

= `-1/6`

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अध्याय 7: Limits - Exercise 7.1 [पृष्ठ १३९]

अध्याय 7 Limits
Exercise 7.1 | Q III. (5) | पृष्ठ १३९

Evaluate the following limit:

`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`

Evaluate the following limit :

`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`

Evaluate the following :

`lim_(x -> 0)[x/(|x| + x^2)]`

Evaluate the following :

Given that 7x ≤ f(x) ≤ 3x² – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`

Evaluate the following :

`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`

In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) sin x/x`

x	– 0.1	– 0.01	– 0.001	0.001	0.01	0.1
f(x)	0.99833	0.99998	0.99999	0.99999	0.99998	0.99833

In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

`lim_(x -> 0) sec x`

In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

`lim_(x -> x/2) tan x`

If the limit of f(x) as x approaches 2 is 4, can you conclude anything about f(2)? Explain reasoning

Evaluate : `lim_(x -> 3) (x^2 - 9)/(x - 3)` if it exists by finding `f(3^-)` and `f(3^+)`