Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`
Advertisements
उत्तर
`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9)) = lim_(x -> 3) ((x + 3)(x - 3))/(x^2(x - 3)2`
= `lim_(x -> 3) (x +3)/(x^2(x - 3))`
To find he left limit
Put x = 3 – h
Where h > 0
When x → 3
We have h → 0
`lim_(x -> 3^-) (x^2 - 9)/(x^2(x^2 - 6x + 9)) = lim_("h" -> 0) (3 - "h" + 3)/((3 - "h")^2 (3 - "h" - 3))`
= `lim_("h" -> 0) (6 - "h")/(-"h"(3- "h")^2`
= `- lim_("h" -> 0) (6 - "h")/("h"(3 - "h")^2`
= `- (6 - 0)/(0(3 - 0)^2`
= `- 6/0`
`lim_(x -> 3^-) (x^2 - 9)/(x^2(x^2 - 6x + 9)) = - oo`
To find he right limit
Put x = 3 + h
Where h > 0
When x → 3
We have h → 0
`lim_(x -> 3^+) (x^2 - 9)/(x^2(x^2 - 6x + 9)) = lim_("h" -> 0) (3 + "h" + 3)/((3 + "h")^2 (3 + "h" - 3))`
= `lim_("h" -> 0) (6 + "h")/("h"(3 + "h")^2`
= `lim_("h" -> 0) (6 + "h")/("h"(3 + "h")^2`
= `(6 + 0)/(0(3 + 0)^2`
= `6/0`
`lim_(x -> 3^+) (x^2 - 9)/(x^2(x^2 - 6x + 9)) = oo`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(x -> 3)[sqrt(2x + 6)/x]`
Evaluate the following :
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`
In problems 1 – 6, using the table estimate the value of the limit.
`lim_(x -> 2) (x - 2)/(x^2 - x - 2)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.344820 | 0.33444 | 0.33344 | 0.333222 | 0.33222 | 0.332258 |
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 2) (x - 2)/(x^2 - 4)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.25641 | 0.25062 | 0.250062 | 0.24993 | 0.24937 | 0.24390 |
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 -> 3) 1/(x - 3)`
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 -> 5) |x - 5|/(x - 5)`
Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.
f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`
Sketch the graph of a function f that satisfies the given value:
f(– 2) = 0
f(2) = 0
`lim_(x -> 2) f(x)` = 0
`lim_(x -> 2) f(x)` does not exist.
Evaluate the following limits:
`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`
Evaluate the following limits:
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`
Evaluate the following limits:
`lim_(x -> oo)(1 + 1/x)^(7x)`
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
Evaluate the following limits:
`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)]`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Evaluate the following limits:
`lim_(x -> ) (sinx(1 - cosx))/x^3`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.
The value of `lim_(x rightarrow 0) (sqrt((1 + x^2)) - sqrt(1 - x^2))/x^2` is ______.
