Advertisements
Advertisements
प्रश्न
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`
Advertisements
उत्तर
`lim_(x -> x/2) tan x`
y = f(x) = sec x
From the graph at x = `pi/2`, the curve does not intersect the line x = `pi/2`
At x = `pi/2`, he value of the function y = f(x) does not exist.
Hence `lim_(x -> x/2) tan x` does not exist.
APPEARS IN
संबंधित प्रश्न
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> -3) (3x + 2)` = – 7
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 -> 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 problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (cos x - 1)/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.0001 | 0.01 | 0.1 |
| f(x) | 0.04995 | 0.0049999 | 0.0004999 | – 0.0004999 | – 0.004999 | – 0.04995 |
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) (4 - x)`
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.
If the limit of f(x) as x approaches 2 is 4, can you conclude anything about f(2)? Explain reasoning
Evaluate the following limits:
`lim_(x -> 2) (x^4 - 16)/(x - 2)`
Evaluate the following limits:
`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`
Evaluate the following limits:
`lim_(x -> oo)(1 + "k"/x)^("m"/x)`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(2) - sqrt(1 + cosx))/(sin^2x)`
Choose the correct alternative:
`lim_(x -> 0) sqrt(1 - cos 2x)/x`
Choose the correct alternative:
If `f(x) = x(- 1)^([1/x])`, x ≤ 0, then the value of `lim_(x -> 0) f(x)` is equal to
Choose the correct alternative:
`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =
`lim_(x -> 0) ((2 + x)^5 - 2)/((2 + x)^3 - 2)` = ______.
`lim_(x -> 5) |x - 5|/(x - 5)` = ______.
`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.
