Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x`
उत्तर
We know `lim_(x -> 0) (sinx)/x` = 1
Sin C – sin D = `2 cos ("C" + "D")/2 * sin ("C" - "D")/2`
`sin("a" + x) - sin("a" - x) = 2 cos(("a" + x + "a" - x)/2) xx sin(("a" + x ("a" - x))/2)`
= `2 cos ((2"a")/2) sin (("a" + x - "a" + x)/2)`
= `2 cos "a" * sin ((2x)/2)`
= 2 cos a sin x
`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x = lim_(x -> 0) (2 cos "a" sin x)/x`
= `2cos "a" lim_(x -> 0) (sinx)/x`
= `2 cos "a" xx 1`
= 2 cos a
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(z -> -3) [sqrt("z" + 6)/"z"]`
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following limit :
`lim_(x -> 1) [(x + x^3 + x^5 + ... + x^(2"n" - 1) - "n")/(x - 1)]`
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 exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`
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 -> 1) sin pi x`
Sketch the graph of a function f that satisfies the given value:
f(0) is undefined
`lim_(x -> 0) f(x)` = 4
f(2) = 6
`lim_(x -> 2) f(x)` = 3
Write a brief description of the meaning of the notation `lim_(x -> 8) f(x)` = 25
Find the left and right limits of f(x) = tan x at x = `pi/2`
Evaluate the following limits:
`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`
Evaluate the following limits:
`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
`lim_(x -> 0) (sin 4x + sin 2x)/(sin5x - sin3x)` = ______.
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.
