0
The limit as x approaches zero of sin(x) over x can be determined using the squeeze theorem.
For 0 < x < pi/2, sin(x) < x < tan(x)
Divide by sin(x), and you get 1 < x/sin(x) < tan(x)/sin(x)
That is the same as 1 < x/sin(x) < 1/cos(x)
But the limit as x approaches zero of 1/cos(x) is 1,
so 1 < x/sin(x) < 1
which means that the limit as x approaches zero of x over sin(x) is 1, and that also means the inverse; the limit as x approaches zero of sin(x) over x is 1.
You can also solve this using deriviatives...
The deriviative d/dxx is 1, at all points. The deriviative d/dxsin(x) at x=0 is also 1.
This means you have the division of two functions, sin(x) and x, at a point where their slope is the same, so the limit reduces to 1 over 1, which is 1.
Still curious? Ask our experts.
Chat with our AI personalities
Lao
Vivi
CoachAdd your answer:
Earn +20 pts
Logic for Fibonacci series in delphi?
i=0 j=0 sum=1 limit = ARGV[0] while sum < limit print sum i = j j = sum sum = i + j end
What speed motor tends to run when field coil opens?
Zero speed [ 0 rpm]
Example problem using nested if statement in C programming?
if (i<0) if (j<0) printf ("i<0 and j<0"); else printf ("What's now? Where this else belongs to? It's i>=0 or i<0 and j>=0?"); The answer is: if (i<0) if (j<0) printf ("i<0 and j<0"); else printf ("i<0 and j>=0"); Of course you can use brackets: if (i<0) { if (j<0) printf ("i<0 and j<0"); else printf ("i<0 and j>=0"); } or if (i<0) { if (j<0) printf ("i<0 and j<0"); } else printf ("i>=0");
How to write a program that for all integers i j k and l from1 to1000 finds and prints all combinations of i j k and l such that i plus j plus k equals l and ijkl?
Some of ways to solve this problem is to use four nested loops for().Something like following:...const int limit = 1000;//some code herefor (int i(0); i
1 22 333 4444 55555 i need a java program that generates a triangle with these numbers?
Implement this method: public static void makeTriangle(int limit) { int count = 0; for(int i = 1; i <= limit; i++) { count = i; while(count > 0) { System.out.print(i); count--; } System.out.println(); } }
