Question

7. (30 points) Evaluate the limit (if it exists). If it is an infinite limit, state whether the limit approaches ? or -?. a) lim_{x??} ln(2+cos x) b) lim_{x?-2^+} rac{|x+2|}{x+2} c) lim_{x?-7^+} rac{2x}{(x+7)^4} d) lim_{x?-4} rac{x^2-16}{x^2+7x+12} e) lim_{x?0} rac{sqrt{x+4}-2}{x} f) lim_{x?3^+} f(x) where f(x)=egin{cases}4x-5, & xle3 \ 2-x^2, & x>3end{cases}

          7. (30 points) Evaluate the limit (if it exists). If it is an infinite limit, state whether the limit approaches ? or -?.
a) lim_{x??} ln(2+cos x)
b) lim_{x?-2^+} rac{|x+2|}{x+2}
c) lim_{x?-7^+} rac{2x}{(x+7)^4}
d) lim_{x?-4} rac{x^2-16}{x^2+7x+12}
e) lim_{x?0} rac{sqrt{x+4}-2}{x}
f) lim_{x?3^+} f(x) where f(x)=egin{cases}4x-5, & xle3 \ 2-x^2, & x>3end{cases}

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7. (30 points) Evaluate the limit (if it exists). If it is an infinite limit, state whether the limit approaches ? or -?.
a) limx?? ln(2+cos x)
b) limx?-2^+ rac|x+2|x+2
c) limx?-7^+ rac2x(x+7)^4
d) limx?-4 racx^2-16x^2+7x+12
e) limx?0 racsqrtx+4-2x
f) limx?3^+ f(x) where f(x)=egincases4x-5, xle3 2-x^2, x>3endcases

Added by Leslie P.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Ahmet Yavas

05/22/2022

Step 1

Plugging in $x = 2$, we get $\ln(2 + \cos 2) = \ln(2 + (-1)) = \ln(1) = 0$. Show more…

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7. (30 points) Evaluate the limit (if it exists). If it is an infinite limit, state whether the limit approaches ∞ or -∞.

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Transcript

00:01 There are given some limit questions.

00:04 So let's start with the first one.

00:06 So it says limit 1x goes to pi.

00:10 The given function is ln2 plus cosine x.

00:13 So the first thing for the limit questions, we have to just plug in the values of pi to the equation, which is ln2 plus cosine pi.

00:23 So the cosine pi is negative 1, ln2 minus 1, which is ln1.

00:27 So the ln1 is equal to, let me just write zero.

00:31 So the limit value for this function, which is zero.

00:34 The next one is, let's say this is part b.

00:37 So the limit when x goes to negative 2 from right, so the given function, which is absolute value of x plus 2 and divided by x plus 2.

00:46 When you're just approaching negative 2 from right, the inside of the absolute value is a positive.

00:52 So i can just write as when x goes to negative 2 from right, that would be plus 2 and x plus 2.

01:00 These are cancelled to each other, so which is left 1.

01:03 So the limit value here, which is 1.

01:07 And for part c, the limit when x goes to negative 7 from right, the given information is 2x, which is x plus 7 to the power 4.

01:17 Let's plug in the value.

01:19 This is 2 times negative 7 divided by negative 7 from right plus 7 to the power 4.

01:25 So the denominator just approaching to the number 0.

01:29 So 0 to the power 4, which is 0.

01:32 So this is negative 14 over 0...

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