Evaluate the following limits(if exits)
\(1.\lim_{x\rightarrow 9}(x(\sqrt{3}-3))/(x-9)\)
\(4.\lim_{x\rightarrow +infinity} (3+\frac{2}{x})(cos\frac{1}{x})\)
\(5.(\sqrt{x^2+x}-\sqrt{x^2-cx}), where c is a positive constant \)
\(3.\lim_{x\rightarrow 0}(1-cos(ax))/({x}^{2})\)
\(2.\lim_{x\rightarrow-infinity}( {x}^{1/3}-5x+3)/(2x+{x}^{2/3}-4)\)
lim x-->9 (x sqrt(3) - 3) / (x - 9)
(two-sided limit does not exist)
lim_(x->9^-) (x sqrt(3)-3)/(x-9) = -infinity (Limit from the left)
lim_(x->9^+) (x sqrt(3)-3)/(x-9) = infinity (limit from the right)
lim x-->∞ (3 + 2/x) * cos(1/x)
Find the following limit:
lim_(x->infinity) cos(1/x) (3+2/x)
Applying the product rule, write lim_(x->infinity) (3+2/x) cos(1/x) as (lim_(x->infinity) (3+2/x)) (lim_(x->infinity) cos(1/x)):
lim_(x->infinity) (3+2/x) lim_(x->infinity) cos(1/x)
Using the fact that cosine is a continuous function, write lim_(x->infinity) cos(1/x) as cos(lim_(x->infinity) 1/x):
lim_(x->infinity) (3+2/x) cos(lim_(x->infinity) 1/x)
Let epsilon>0. Then for all x > 1/epsilon, 1/x<1/(1/epsilon) = epsilon, so
lim_(x->infinity) 1/x = 0:
cos(0) lim_(x->infinity) (3+2/x)
cos(0) = 1:
lim_(x->infinity) (3+2/x)
3+2/x = lim_(x->infinity) 3+2 (lim_(x->infinity) 1/x):
lim_(x->infinity) 3+2 lim_(x->infinity) 1/x
Let epsilon>0. Then for all x > 1/epsilon, 1/x<1/(1/epsilon) = epsilon, so
lim_(x->infinity) 1/x = 0:
lim_(x->infinity) 3+2×0
Since 3 is constant, lim_(x->infinity) 3 = 3:
3+2×0
3+2 0 = 3:
Answer: |=3
lim x-->0 ((1 - cos(ax)) / x^2)
Find the following limit: lim_(x->0) (1-cos(a x))/x^2 Applying l'Hôpital's rule, we get that lim_(x->0) (1-cos(a x))/x^2 = lim_(x->0) ( d/( dx)(1-cos(a x)))/( d/( dx) x^2) = lim_(x->0) (a sin(a x))/(2 x) lim_(x->0) (a sin(a x))/(2 x) (a lim_(x->0) (sin(a x))/x)/(2) Applying l'Hôpital's rule, we get that lim_(x->0) (sin(a x))/x | = | lim_(x->0) ( d/( dx) sin(a x))/(( dx)/( dx)) | = | lim_(x->0) (a cos(a x))/1 | = | lim_(x->0) a cos(a x) (a lim_(x->0) a cos(a x))/2 lim_(x->0) a cos(a x) = a cos(0 a) = a: (a a)/2 1/2 a a = a^2/2: Answer: | | a^2/2
lim x-->∞ (x^1/3 - 5x + 3) / (2x + x^2/3 - 4)
Find the following limit: lim_(x->infinity) (3+x^(1/3)-5 x)/(x^2/3+2 x-4) (3+x^(1/3)-5 x)/(x^2/3+2 x-4) = (3+x^(1/3)-5 x)/(x^2/3+2 x-4): lim_(x->infinity) (3+x^(1/3)-5 x)/(x^2/3+2 x-4) The leading term in the denominator of (3+x^(1/3)-5 x)/(x^2/3+2 x-4) is x^2. Divide the numerator and denominator by this: lim_(x->infinity) (3/x^2+1/x^(5/3)-5/x)/(1/3+2/x-4/x^2) The expressions 3/x^2, 1/x^(5/3), -5/x, -4/x^2 and 2/x all tend to zero as x approaches infinity: 0/(1/3) 0/(1/3) = 0: Answer: | | 0
lim x-->∞ (sqrt(x^2 + x)) - sqrt(x^2 - cx), c=positive constant
You didn't specify the limit:lim_(x->infinity) (sqrt(x^2+x)-sqrt(x^2-c x)) = (1+c)/2
