why would the upper range be "≤ 2√x"??
why would the upper range be "≤ 2√x"??
I was unable to find a method online how to use the δ-ε definition of a limit to prove that the limit is L for a function which includes an absolute value. My function is ...
lim ...
find lim x --- -1 6x+5 _______ ...
lim cosθ-(1/√2)/(θ-π/4) θ→π/4 I rewrote θ as h so I have lim which is cosh-(1/√2)/h-(π/4) ...
Find the values of a and b that make the function f(x) continuous if: f(x)=sinx/x for x less than zero ax+b for x greater than or equal to zero and less than or equal to two x^2+3...
f(x)= (x^2-144)/ (x-12) x≠a k x=a
lim of (1-cos5x)/(cos6x-1) as x->0
fin the limit of (1-cos2x-cos4x)/x as x->0 and show all work!
lim (x+1)/(x+2) x->3+ how would a + or - on c change anything when solving without a graph?
no graph provided lim x=? x->9
If limit as x->3 of (f(x)-4)/(x-3)=3, find the limit as x->3 of f(x). If limit as x->3 of (f(x)-4)/(x-3)=6, find the limit as x->3 of f(x).
Find the x-value-(s) for which the following function is discontinuous. For each discontinuity state the type as either removable or non-removable. Please show some of the steps.
Hi, buddies. Please me in my task to answer this questions. As far as my questions, I'm just looking for "What are the real world application of limits (calculus limits) for Industrial Engineering...
Given that lim x->a f(x)=0 lim x->a g(x)=0 lim x->a h(x)=1 lim x->a p(x)=infinity lim x->a q(x)= infinity. THEN (a) lim x->a f(x)/g(x)= (b)...
Let f(x) = √(-3-x)+1, if x<-4 let f (x)= 1, if x=-4 let f (x)= 3x+14, if x>-4 Calculate the following limits. Enter DNE for a limit which does not exist lim x->-4- f(x)= lim...
the limit as x approaches 0 1 - 1 ...
the limit as x approaches infinity 4x3-5x/8x43x2-2
find the lim as x approches 5 of f(x)=In(x+3) find the lim 2/x+3 x-->3 what...
Lim h-->0 (((.5 + h)5) - ((.5)5))/h