Author Archives: Engineering Math

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 8

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PROBLEM:

Evaluate lim⁡x→8 x3−2x−8\displaystyle \lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}.

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SOLUTION:

A straight substitution of x=8x=8 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→8 x3−2x−8=lim⁡x→ 8 x3−2x−8⋅x23+2x3+4x23+2x3+4=lim⁡x→8 x−8(x−8)(x23+2x3+4)=lim⁡x→8 1(x23+2x3+4)=1(823+283+4)=14+4+4=112  (Answer)\begin{align*} \\ \lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}& =\lim\limits_{x\to \:8}\:\frac{\sqrt[3]{x}-2}{x-8}\cdot \frac{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}\\ \\ & =\lim\limits_{x\to 8}\:\frac{x-8}{\left(x-8\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\ \\ & =\lim\limits_{x\to 8}\:\frac{1}{\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\ \\ & =\frac{1}{\left(\sqrt[3]{8^2}+2\sqrt[3]{8}+4\right)}\\ \\ & =\frac{1}{4+4+4}\\ \\ & =\frac{1}{12} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 7

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PROBLEM:

Evaluate lim⁡x→1 x−1x+3−2 \displaystyle \lim\limits_{x\to 1}\:\frac{x-1}{\sqrt{x+3}-2}.

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 SOLUTION:

A straight substitution of  x=1x=1 leads to the indeterminate form 00 \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→1 x−1x+3−2=lim⁡x→ 1 x−1x+3−2⋅x+3+2x+3+2=lim⁡x→1(x−1)(x+3+2)(x+3)−22=lim⁡x→1(x−1)(x+3+2)x−1=lim⁡x→1x+3+2=1+3+2=4+2=2+2=4  (Answer)\begin{align*} \\ \lim\limits_{x\to 1}\:\frac{x-1}{\sqrt{x+3}-2}& =\lim\limits_{x\to \:1}\:\frac{x-1}{\sqrt{x+3}-2}\cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\\ \\ & =\lim\limits_{x\to 1}\frac{\left(x-1\right)\left(\sqrt{x+3}+2\right)}{\left(x+3\right)-2^2}\\ \\ & =\lim\limits_{x\to 1}\frac{\left(x-1\right)\left(\sqrt{x+3}+2\right)}{x-1}\\ \\ & =\lim\limits_{x\to 1}\sqrt{x+3}+2&\\ \\ & =\sqrt{1+3}+2\\ \\ &=\sqrt{4}+2\\ \\ & =2+2\\ \\ & =4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 6

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PROBLEM:

Evaluate lim⁡x→0 x+16−4x \displaystyle\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}.

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 SOLUTION:

A straight substitution of x=0x=0 leads to the indeterminate form 00 \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows.

lim⁡x→0 x+16−4x=lim⁡x→0 x+16−4x⋅x+16+4x+16+4=lim⁡x→0 (x+16)−42x(x+16+4)=lim⁡x→0 x+16−16x(x+16+4)=lim⁡x→0 xx(x+16+4)=lim⁡x→0 1x+16+4= 10+16+4= 14+4= 18  (Answer)\begin{align*} \\ \lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x} & =\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}\cdot \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}\\ \\ & =\lim\limits_{x\to 0}\:\frac{\left(x+16\right)-4^2}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x+16-16}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{1}{\sqrt{x+16}+4}\\ \\ & =\:\frac{1}{\sqrt{0+16}+4}\\ \\ & =\:\frac{1}{4+4}\\ \\ & =\:\frac{1}{8} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}

 

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College Physics by Openstax Chapter 2 Problem 41

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y0=0.

Solution:

The given known quantities are:a=−9.8 m/s2a=-9.8\:\text{m/s}^2; yo=0 my_o=0\:\text{m}; and voy=+15 m/sv_{oy}=+15\:\text{m/s}.

To compute for the displacement, we use the formula

Δy=voyt+12at2\Delta y=v_{oy}t+\frac{1}{2}at^2

and to compute for the final velocity, we use the formula

vfy=voy+atv_{fy}=v_{oy}+at

Part A

The displacement at t=0.500 st=0.500 \ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(0.500 s)+12(−9.8 m/s2)(0.500 s)2Δy=6.28 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\ \Delta y & =6.28\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=0.500 st=0.500 \ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(0.500 s)vfy=10.1 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\ v_{fy} & =10.1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The displacement at t=1.000 st=1.000 \ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(1.000 s)+12(−9.8 m/s2)(1.000 s)2Δy=10.1 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right)^2 \\ \Delta y & =10.1\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.000 st=1.000\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(1.000 s)vfy=5.20 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right) \\ v_{fy} & =5.20\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The displacement at t=1.500 st=1.500\ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(1.500 s)+12(−9.8 m/s2)(1.500 s)2Δy=11.5 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right)^2 \\ \Delta y & =11.5\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.500 st=1.500\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(1.500 s)vfy=0.300 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right) \\ v_{fy} & =0.300\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The displacement at t=2.000 st=2.000\ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(2.000 s)+12(−9.8 m/s2)(2.000 s)2Δy=10.4 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(2.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right)^2 \\ \Delta y & =10.4\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=2.000 st=2.000\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(2.000 s)vfy=−4.600 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right) \\ v_{fy} & =-4.600\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 40

(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration.

(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

Solution:

Part A

There are two parts to the race and must be treated separately since acceleration is not uniform over the race. We will divide the race into Δx1\Delta x_1 (while accelerating) and Δx2\Delta x_2 (with constant speed), where Δx1+Δx2=100 m\Delta x_1 + \Delta x_2 = 100 \ \text{m} .

For Δx1\Delta x_1:

During the accelerating period, we are going to use the formula Δx=v0t+12at2\Delta x=v_0t+\frac{1}{2}at^2, since we know that a=Δvt=vmax−v0t=vmaxt\displaystyle a=\frac{\Delta v}{t}=\frac{v_{max}-v_0}{t}=\frac{v_{max}}{t}; and t=3.00 st=3.00 \ \text{s}.

Δx=v0t+12at2Δx1=0+12at2Δx1=12at2Δx1=12(vmaxt)t2Δx1=12(vmax)tΔx1=12(vmax)(3.00 s)Δx1=1.5vmax\begin{align*} \Delta x & =v_0t+\frac{1}{2}at^2 \\ \Delta x_1 & =0+\frac{1}{2}at^2 \\ \Delta x_1 & =\frac{1}{2}at^2 \\ \Delta x_1 & =\frac{1}{2}\left(\frac{v_{max}}{t}\right)t^2 \\ \Delta x_1 & =\frac{1}{2}\left(v_{max}\right)t \\ \Delta x _1&=\frac{1}{2}\left(v_{max}\right)\left(3.00\:\text{s}\right) \\ \Delta x _1&=1.5v_{max} \end{align*}

When the speed is constant, t=6.69 st=6.69 \ \text{s}, so

Δx2=vmaxtΔx2=vmax(6.69 s)Δx2=6.69vmax\begin{align*} \Delta x_2 & = v_{max}t \\ \Delta x_2 & = v_{max}\left(6.69\:\text{s}\right) \\ \Delta x_2 & =6.69v_{max} \end{align*}

Plugging-in the two equations in the equation Δx1+Δx2=100 m\Delta x_1 + \Delta x_2 = 100 \ \text{m} .

Δx1+Δx2=100 m1.5vmax+6.69vmax=100 m8.19 vmax=100vmax=1008.19vmax=12.2 m/s  (Answer)\begin{align*} \Delta x_1 + \Delta x_2 & = 100 \ \text{m} \\ 1.5v_{max} + 6.69v_{max} & =100 \ \text{m} \\ 8.19\:v_{max} & =100 \\ v_{max} & =\frac{100}{8.19} \\ v_{max} & =12.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, his acceleration can be computed using the formula

a=vmaxta=\frac{v_{max}}{t}

Plugging in the given values

a=vmaxta=12.2 m/s3.00 sa=4.07 m/s2  (Answer)\begin{align*} a & =\frac{v_{max}}{t} \\ a & = \frac{12.2\:\text{m/s}}{3.00\:\text{s}} \\ a & = 4.07\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Similar to part (a), we can plug in the different values for time and total distance:

Δx1+Δx2=2001.5 vmax+(19.30−3.00)vmax=2001.5 vmax+16.30vmax=20017.80vmax=200vmax=20017.80vmax=11.2 m/s  (Answer)\begin{align*} \Delta x_1+ \Delta x_2 & =200 \\ 1.5\:v_{max}+\left(19.30-3.00\right)v_{max} & =200 \\ 1.5\:v_{max}+16.30v_{max} & =200 \\ 17.80v_{max} & =200 \\ v_{max} & =\frac{200}{17.80} \\ v_{max} & = 11.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 5

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PROBLEM:

Evaluate lim⁡x→0 (x+3)2−92x\displaystyle \lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x}.

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 SOLUTION:

A straight substitution of x=0x=0 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→0 (x+3)2−92x= lim⁡x→0 (x+3)2−(3)22x= lim⁡x→0 (x+3−3)(x+3+3)2x=lim⁡x→0 x(x+6)2x=lim⁡x→0 x+62=0+62=62=3  (Answer)\begin{align*} \lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x} & =\:\lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-\left(3\right)^2}{2x}\\ \\ & =\:\lim\limits_{x\to 0}\:\frac{\left(x+3-3\right)\left(x+3+3\right)}{2x}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x\left(x+6\right)}{2x}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x+6}{2}\\ \\ & =\frac{0+6}{2}\\ \\ & =\frac{6}{2}\\ \\ & =3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 4

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PROBLEM:

Evaluate lim⁡x→2(x3−x2−x−22x3−5x2+5x−6) \displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right).

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 SOLUTION:

A straight substitution of x=2x=2 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→2(x3−x2−x−22x3−5x2+5x−6)=lim⁡x→2((x−2)(x2+x+1)(x−2)(2x2−x+3))=lim⁡x→2(x2+x+12x2−x+3)=22+2+12(2)2−2+3=4+2+18−2+3=79  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right)&=\lim\limits_{x\to 2}\left(\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(2x^2-x+3\right)}\right)\\ \\ & =\lim\limits_{x\to 2}\left(\frac{x^2+x+1}{2x^2-x+3}\right)\\ \\ & =\frac{2^2+2+1}{2\left(2\right)^2-2+3}\\ \\ & =\frac{4+2+1}{8-2+3}\\ \\ & =\frac{7}{9} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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College Physics by Openstax Chapter 2 Problem 39

In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

Solution:

There are two parts to the race: an acceleration part and a constant speed part.

For the acceleration part:

We are given the following values: v0=0 mphv_0=0 \ \text{mph} ; vf=60 mph v_f=60 \ \text{mph}; and Δt=4.00 s\Delta t=4.00 \ \text{s} .

First, we need to determine how long (both in distance and time) it takes the motorcycle to finish accelerating. During acceleration, the value of the acceleration is given by

a=60 mph4 sa=\frac{60\:\text{mph}}{4\:\text{s}}

To compute for the time it takes to reach its maximum velocity, we are going to use the formula

vf=v0+atv_f=v_0+at

Solving for time tt in terms of the other variables

t=vf−v0at=\frac{v_f-v_0}{a}

Substituting the given values to solve for t1t_1, the time it takes to accelerate from rest to maximum velocity:

t1=vf−v0at1=183 mph−0 mph(60 mph4 s)t1=12.2 s\begin{align*} t_1 & =\frac{v_f-v_0}{a} \\ t_1 & =\frac{183\:\text{mph}-0\:\text{mph}}{\left(\frac{60\:\text{mph}}{4\:\text{s}}\right)} \\ t_1 & =12.2\:\text{s} \end{align*}

Since we have a constant acceleration, the distance traveled Δx1\Delta x_1 during this period is computed using the formula

Δx1=vavetΔx1=(vf+v02)t\begin{align*} \Delta x_1 & =v_{ave}t \\ \Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\ \end{align*}

Substituting the given values:

Δx1=(vf+v02)tΔx1=(183 mph+0 mph2)(12.2 s)Δx1=(91.5 mph)(1 hr3600 s)(12.2 s)Δx1=0.31 mi\begin{align*} \Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\ \Delta x_1 & =\left(\frac{183\:\text{mph}+0\:\text{mph}}{2}\right)\left(12.2\:\text{s}\right) \\ \Delta x_1 & =\left(91.5\:\text{mph}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right)\left(12.2\:\text{s}\right) \\ \Delta x_1 & =0.31\:\text{mi} \end{align*}

For the constant speed part:

For the next part of the motion, the speed is constant.

We are given the following values: Δx2=5.0 mi−0.3 mi=4.7 mi\Delta x_2=5.0\:\text{mi}-0.3\:\text{mi}=4.7\:\text{mi} .

We are going to solve t2t_2, the time spent on the course at max speed using the formula

Δx2=vmaxt2\Delta x_2=v_{max}t_2

Solving for t2t_2 in terms of the other variables:

t2=Δx2vmaxt_2=\frac{\Delta x_2}{v_{max}}

Substituting the given values:

t2=Δx2vmaxt2=4.7 mi183 mpht2=(0.026 h)(3600 s1 h)t2=92 s\begin{align*} t_2 & =\frac{\Delta x_2}{v_{max}} \\ t_2 & =\frac{4.7\:\text{mi}}{183\:\text{mph}} \\ t_2 & =\left(0.026\:\text{h}\right)\left(\frac{3600\:\text{s}}{1\:\text{h}}\right) \\ t_2 &=92\:\text{s} \end{align*}

For the whole course:

So, the total time is

ttotal=t1+t2ttotal=12.2 s+ 92 sttotal=104 s  (Answer)\begin{align*} t_{total}&=t_1+t_2 \\ t_{total}& =12.2\:\text{s}+\:92\:\text{s} \\ t_{total}& =104\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 3

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PROBLEM:

Evaluate lim⁡x→3(x3−13x+12x3−14x+15) \displaystyle \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right).

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SOLUTION:

A straight substitution of x=3x=3 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows.

lim⁡x→3(x3−13x+12x3−14x+15)=lim⁡x→3((x−3)(x2+3x−4)(x−3)(x2+3x−5))=lim⁡x→3(x2+3x−4x2+3x−5)=(3)2+3(3)−4(3)2+3(3)−5=9+9−49+9−5=1413  (Answer)\begin{align*} \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right)& =\lim\limits_{x\to 3}\left(\frac{\left(x-3\right)\left(x^2+3x-4\right)}{\left(x-3\right)\left(x^2+3x-5\right)}\right)\\ \\ & =\lim\limits_{x\to 3}\left(\frac{x^2+3x-4}{x^2+3x-5}\right)\\ \\ &=\frac{\left(3\right)^2+3\left(3\right)-4}{\left(3\right)^2+3\left(3\right)-5}\\ \\ & =\frac{9+9-4}{9+9-5}\\ \\ & =\frac{14}{13} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 2

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PROBLEM:

Evaluate lim⁡x→2(x2+2x−83x−6)\displaystyle \lim\limits_{x\to 2}\left(\frac{x^2+2x-8}{3x-6}\right)

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SOLUTION:

A straight substitution of x=2x=2 leads to the indeterminate form 00\frac{0}{0}  which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→2(x2+2x−83x−6)=lim⁡x→2((x+4)(x−2)3(x−2))=lim⁡x→2(x+43)=2+43=63=2  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(\frac{x^2+2x-8}{3x-6}\right)& =\lim\limits_{x\to 2}\left(\frac{\left(x+4\right)\left(x-2\right)}{3\left(x-2\right)}\right)\\ \\ &=\lim\limits_{x\to 2}\left(\frac{x+4}{3}\right)\\ \\ &=\frac{2+4}{3}\\ \\ &=\frac{6}{3}\\ \\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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