Here are the solutions to the problems.
1.
Step 1: Substitute x=1 into the expression.
1−11−1=00
This is an indeterminate form.
Step 2: Simplify the expression.
For x=1, the expression x−1x−1 simplifies to 1.
Step 3: Evaluate the limit.
limx→1x−1x−1=limx→11=1
The limit is 1.
2.
Step 1: Substitute x=1 into the expression.
12−112−1=1−11−1=00
This is an indeterminate form.
Step 2: Simplify the expression.
For x=1, the expression x2−1x2−1 simplifies to 1.
Step 3: Evaluate the limit.
limx→1x2−1x2−1=limx→11=1
The limit is 1.
3.
Step 1: Substitute x=3 into the expression.
3−933−27=−627−27=−60=0
This is not an indeterminate form.
Step 2: Evaluate the limit.
limx→3x−9x3−27=0
The limit is 0.
4.
Step 1: Substitute x=0 into the expression.
0(3+0)2−9=032−9=09−9=00
This is an indeterminate form.
Step 2: Expand the numerator.
(3+x)2−9=(9+6x+x2)−9=6x+x2
Step 3: Simplify the expression.
x6x+x2=xx(6+x)=6+xforx=0
Step 4: Evaluate the limit.
limx→0(6+x)=6+0=6
The limit is 6.
5.
Step 1: Substitute x=1 into the expression.
12−3(1)+213−1=1−3+21−1=00
This is an indeterminate form.
Step 2: Factor the numerator and denominator.
The numerator is a difference of cubes: x3−1=(x−1)(x2+x+1).
The denominator is a quadratic: x2−3x+2=(x−1)(x−2).
Step 3: Simplify the expression.
(x−1)(x−2)(x−1)(x2+x+1)=x−2x2+x+1forx=1
Step 4: Evaluate the limit.
limx→1x−2x2+x+1=1−212+1+1=−13=−3
The limit is −3.
6.
Step 1: Substitute x=5 into the expression.
52−2552−7(5)+10=25−2525−35+10=00
This is an indeterminate form.
Step 2: Factor the numerator and denominator.
The numerator is a quadratic: x2−7x+10=(x−2)(x−5).
The denominator is a difference of squares: x2−25=(x−5)(x+5).
Step 3: Simplify the expression.
(x−5)(x+5)(x−2)(x−5)=x+5x−2forx=5
Step 4: Evaluate the limit.
limx→5x+5x−2=5+55−2=103
The limit is 103.
For questions 7-11, we will find the derivative dxdy.
7.
Step 1: Rewrite the function.
y=cosx1=secx
Step 2: Differentiate y with respect to x.
dxdy=dxd(secx)=secxtanx
The derivative is secxtanx.
8.
Step 1: Rewrite the function.
y=sinx1=cscx
Step 2: Differentiate y with respect to x.
dxdy=dxd(cscx)=−cscxcotx
The derivative is −cscxcotx.
9.
Step 1: Apply the chain rule. Let u=x2+4. Then y=u4.
dxdy=dudy⋅dxdu
Step 2: Find dudy and dxdu.
dudy=dud(u4)=4u3
dxdu=dxd(x2+4)=2x
Step 3: Substitute back u=x2+4.
dxdy=4(x2+4)3⋅(2x)=8x(x2+4)3
The derivative is 8x(x2+4)3.
10.
Step 1: Differentiate y with respect to x.
dxdy=dxd(sinx)=cosx
The derivative is cosx.
11.
Step 1: Apply the chain rule. Let u=x2+4. Then y=u3.
dxdy=dudy⋅dxdu
Step 2: Find dudy and dxdu.
dudy=dud(u3)=3u2
dxdu=dxd(x2+4)=2x
Step 3: Substitute back u=x2+4.
dxdy=3(x2+4)2⋅(2x)=6x(x2+4)2
The derivative is 6x(x2+4)2.
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