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Featured 2 months ago

**1.**

Letting,

**2.**

We subst.

**3.**

Taking

**Enjoy Maths., and Spread the Joy!**

Featured 2 months ago

Without L'Hospital's rule see below.

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

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

As

Featured 1 month ago

Please see below.

We need to find

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

We will use the following limits:

And

To evaluate the limit:

Featured 1 month ago

Evaluate the limit:

Note that:

so we can simplify:

Multiply numerator and denominator by

Now consider the limit:

Substituting

that is a well known trigonometric limit:

So:

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

Featured 1 month ago

Use the substitution

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

#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

#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#

Featured 3 weeks ago

The limit is

We can rewrite as

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

We see that

#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

#L = 2sec^2(0)#

#L = 2(1)#

#L = 2#

A graphical verification confirms.

In the above graph, the red curve is

Hopefully this helps!

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