Let lim x-->-6 f(x)=6 and lim x-->-6 f(x)=3.?
There's a limit property which states the limit of the sum of two functions is the sum of their limits. It is the same for the difference because lim a.f(x) = a.lim f(x) is true ( "a" is a real constant ). lim [ f(x) - g(x) ] = lim f(x) - lim g(x) In this problem lim x-->-6 [ f(x) - g(x) ] = lim x-->-6 f(x) - lim x-->-6 g(x) Since lim x-->-6 f(x) = 6 and lim x-->-6 g(x) = 3 lim x-->-6 [ f(x) - g(x) ] = 3 So, the correct choice is A.
If lim x-> a [f(x) + g(x)] = 2, and lim x-> a [f(x) - g(x)] = 1, find lim x-> a [f(x)*g(x)]?
So you have this lim x -> a [f(x) + g(x)] = 2 and lim x -> a [f(x) - g(x)] = 1 With that you can transform it into this system of equations lim [x->a] f(x) + lim[x->a] g(x) = 2 lim [x->a] f(x) - lim[x->a] g(x) = 1 Solve the second equation for lim [x->a] f(x) to get lim [x->a] f(x) = 1 + lim[x->a] g(x) Substitute in the first equation to get 1 + lim[x->a] g(x) + lim[x->a] g(x) = 2 2 lim[x->a] g(x) = 1 so lim[x->a] g(x) = 1/2 and using lim [x->a] f(x) = 1 + lim[x->a] g(x) we get lim [x->a] f(x) = 3/2 So to find this lim x -> a [f(x)*g(x)] we get lim x -> a [f(x)*g(x)] = lim [x->a] f(x) * lim [x->a] g(x) = 3/2 * 1/2 = 3/4 so lim x -> a [f(x)*g(x)] = 3/4
Lim h->0 f(x+h)-f(x)/h when f(x)=3√x?
f(x) = 3 (x)^1/2 f(x+h) = 3 (x+h)^1/2 lim h-->0 [3(x+h)^1/2 - 3 (x)^1/2] / h multiply top and bottom by [(x+h)^1/2 + (x)^1/2] this will get rid of the square roots in the numerator and give you a chance to eliminate h. lim h-->0 [3(x+h) - 3 (x)] / h[(x+h)^1/2 + (x)^1/2] lim h-->0 3h / h [(x+h)^1/2 + (x)^1/2] now as we can cancel the h from the numerator and the denominator. lim h-->0 3 / [(x+h)^1/2 + (x)^1/2] and as h goes to 0 it becomes insignificant where it is. = 3 / (2 (x)^1/2)
Consider the function belowf(x)=1 if x is rational = -1 if x is irrationallim|f(x)| is clearly 1. How abt limf(x)? Not defined. So, it is always not necessary to have the equality.
Find lim if 2x + 2 ≤ f(x) ≤ x2 + 3. x→1?
2x + 2 ≤ f(x) ≤ x^2 + 3 and lim x→1, 2x + 2 = lim x→1, x^2 + 3 = 4 => we can apply squeeze theorem : lim x→1, f(1) = 4 4 ≤ f(x) ≤ 4
Lim(f(x) + g(x)) as x--> -5 when...?
as x -->5 then lm f(x) --> - 3 and g(x)-->1 so as x-->- 5 then Lim [f(x) + g(x)] --> - 3 + 1 = - 2
No.The easiest examples are at asymptotes. Consider [math]f(x)=\frac{1}{x}[/math] and [math]g(x)=tan(x)[/math]. Next, let’s consider the example of [math]n=\frac{\pi}{2}.[/math]The value of [math]lim_{x \to \frac{\pi}{2}} tan(x)[/math] doesn’t exist because it can be [math]+\infty[/math] or [math]-\infty[/math] depending whether it’s being approached from the positive or negative side.But when the denominator of a fraction approaches either infinity while the numerator stays constant, the fraction tends to 0. Thus, [math]lim_{x \to \frac{\pi}{2}} \frac{1}{tan(x)}[/math] evaluates to 0.
There are many . as an example f(x)=1-e^(-x) it's limit is 1 as x->infinityIn fact for every function g(x) which tends to zero as x->infinity you can definef(x)=a+g(x)Where a is a constant. As g(x)->0f(x)->aAnd for every function g(x) which become infinity as x->infinity you can define f(x)=a+1/g(x)As g(x) ->infinityf(x) ->a.
Only if the function is continuous at point x. This is actually the definition of a function continuous at x.The definition is:We say that function [math]f[/math] is continuous at point [math]x_{0}[/math] if (and only if) [math]\large \lim_{x \rightarrow x_{0}}f(x)=f(x_{0})[/math].You can't actually prove a definition. There is nothing to prove there, definition is just naming something. The definiedum is relatively long, so it's much easier to just say continuous function instead of dragging that with you every time you need to explain this type of function. But, if your question had a hidden assumption that the function is continuous then the proof is just calling upon the definition, nothing else.*The only if part is in the brackets because, by agreement, it's assumed that the definition is an equivalence relation, but I've put it anyways just to make it more clear.With the detail given above I have to edit my answer.As I can see the staring assumption was that the function has a derivative. And going from there you've given the very nice (and correct) proof that a function having derivative is enough of a condition for it to be continuous. If you came to this proof all by yourself very nice, and good job! :)