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Answer:

#1. ln|(x+3)+sqrt(x^2+6x+10)|+C_1#.

#2. sin^-1((x-3)/3)+C_2#.

#3. tan^-1(2x+3)+C_3#.

Explanation:

1. #intdx/sqrt(x^2+6x+10)=I_1," say."#

#:. I_1=intdx/sqrt{(x+3)^2+1}#.

Letting, #(x+3)=u, dx=du#.

#:. I_1=int(du)/sqrt(u^2+1^2)#,

#=ln|(u+sqrt(u^2+1^2))|#,

#rArr I_1=ln|(x+3)+sqrt(x^2+6x+10)|+C_1#.

2. #intdx/sqrt(6x-x^2)=I_2," say."#

#:. I_2=intdx/sqrt(9-9+6x-x^2)=intdx/sqrt{3^2-(x-3)^2}#.

We subst. #x-3=y," so that, "dx=dy#.

#:. I_2=intdy/sqrt(3^2-y^2)#,

#=sin^-1(u/3)#.

#rArr I_2=sin^-1((x-3)/3)+C_2#.

3. #intdx/(2x^2+6x+5)=I_3," say"#.

#:. I_3=intdx/{2(x^2+3x+5/2)#,

#=1/2intdx/(x^2+3x+9/4+1/4)#,

#=1/2intdx/{(x+3/2)^2+(1/2)^2#.

Taking #(x+3/2)=t," we have, "dx=dt#.

#:. I_3=1/2intdt/(t^2+(1/2)^2#,

#=1/2{1/(1/2)tan^-1(t/(1/2))}#,

#=tan^-1(2t)#.

#rArr I_3=tan^-1(2(x+3/2)=tan^-1(2x+3)+C_3#.

Enjoy Maths., and Spread the Joy!

Answer:

Without L'Hospital's rule see below.

Explanation:

#cos A = sin (pi/2-A)# so we have

#cos(pi/x)/(x-2) = sin(pi/2-pi/x)/(x-2)#

# = sin((pi(x-2))/(2x))/(x-2)#

# = pi/(2x) (sin((pi(x-2))/(2x)))/((pi(x-2))/(2x)#

As #xrarr2#, we have #((pi(x-2))/(2x))rarr0# so we get

#lim_(xrarr2)pi/(2x) (sin((pi(x-2))/(2x)))/((pi(x-2))/(2x)) = pi/(2(2)) (1) = pi/4#

Answer:

Please see below.

Explanation:

#f(x) = cos(x^2-1)#

We need to find

#lim_(hrarr0)(f(x+h)-f(x))/h = lim_(hrarr0)(cos((x+h)^2-1)-cos(x^2-1))/h#

Let's focus on the expression whose limit we need.

#(cos((x^2-1)+(2xh+h^2))-cos(x^2-1))/h#

# = (cos(x^2-1)cos(2xh+h^2) - sin(x^2-1)sin(2xh+h^2)-cos(x^2-1))/h#

# = cos(x^2-1)(cos(2xh+h^2)-1)/h - sin(x^2-1)sin(2xh+h^2)/h#

# = cos(x^2-1)(cos(2xh+h^2)-1)/(h(2x+h))(2x+h) - sin(x^2-1)sin(2xh+h^2)/(h(2x+h))(2x+h)#

We will use the following limits:

#lim_(hrarr0)(cos(2xh+h^2)-1)/(h(2x+h)) = lim_(trarr0)(cost-1)/t = 0#

#lim_(hrarr0)sin(2xh+h^2)/(h(2x+h)) = lim_(trarr0)sint/t = 1#

And #lim_(hrarr0) (2x+h) = 2x#

To evaluate the limit:

#cos(x^2-1) ( 0) (2x) - sin(x^2-1) * (1) * (2x) = -2xsin(x^2-1)#

Answer:

#lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2 = 8#

Explanation:

Evaluate the limit:

#lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2#

Note that:

#(x^2-4) = (x+2)(x-2)#

so we can simplify:

#lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2 = lim_(x->2) ((x+2)(x-2)tan(2x-4))/(x-2)^2 = lim_(x->2) ((x+2)tan(2x-4))/(x-2)#

Multiply numerator and denominator by #2#:

#lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2 = lim_(x->2) (2(x+2)tan(2x-4))/(2x-4)#

Now consider the limit:

#lim_(x->2) tan(2x-4)/(2x-4)#

Substituting #y=2x-4#, as #lim_(x->2) 2x-4 = 0# we have:

#lim_(x->2) tan(2x-4)/(2x-4) = lim_(y->0) tany/y#

that is a well known trigonometric limit:

#lim_(y->0) tany/y = lim_(y->0) 1/cosy siny/y = 1#

So:

#lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2 = lim_(x->2) 2(x+2) * lim_(x->2) tan(2x-4)/(2x-4) = 8*1 = 8#

graph{((x^2-4)tan(2x-4))/(x-2)^2 [-4.873, 5.127, 6.26, 11.26]}

Answer:

Use the substitution #sqrtx+1=sqrt7sectheta#.

Explanation:

Let

#I=int1/sqrt(4x+8sqrtx-24)dx#

Complete the square in the square root:

#I=1/2int1/sqrt((sqrtx+1)^2-7)dx#

Apply the substitution #sqrtx+1=sqrt7u#:

#I=int(sqrt7u-1)/sqrt(u^2-1)du#

Rearrange:

#I=sqrt7intu/sqrt(u^2-1)du-int1/sqrt(u^2-1)du#

Apply the substitution #u=sectheta#:

#I=sqrt7sqrt(u^2-1)-intsecthetad theta#

Hence

#I=sqrt(7u^2-7)-ln|sectheta+tantheta|+C#

Reverse the last substitution:

#I=sqrt(7u^2-7)-ln|u+sqrt(u^2-1)|+C#

Reverse the first substitution:

#I=sqrt((sqrtx+1)^2-7)-ln|(sqrtx+1)+sqrt((sqrtx+1)^2-7)|+C#

Simplify:

#I=sqrt(x^2+2sqrtx-6)-ln|(sqrtx+1)+sqrt(x^2+2sqrtx-6)|+C#

Evaluate the Limit?

Noah G
Noah G
Featured 3 weeks ago

Answer:

The limit is #2#

Explanation:

We can rewrite as

#L = lim_(x-> oo) tan(2/x)/(1/x)#

We see that #tan(2/x)# converges to #0# as #x-> oo# because the larger the value of #x# the smaller the expression within the tangent becomes and the closer #tan(2/x)# becomes to #0#. Also, the limit #lim_(x-> oo) 1/x = 0# is commonly used. Therefore, we may use l;Hosptial's rule.

#L = lim_(x->oo) (-2/x^2 * sec^2(2/x))/(-1/x^2)#

#L = lim_(x-> oo) 2sec^2(2/x)#

The same principal applies with #2/x# as #1/x#: the limit as #x# approaches infinity always remains #0#.

#L = 2sec^2(0)#

#L = 2(1)#

#L = 2#

A graphical verification confirms.

enter image source here

In the above graph, the red curve is #xtan(2/x)# and the blue line is #y = 2#. As you can see, the curve converges onto the line.

Hopefully this helps!

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