WORKSHEET L.1-3 Finding Limits Analytically Common | Chegg.com - Free Printable

Let’s go through each problem one by one. We’ll plug in the value that x is approaching, and see if we get a valid number. If we get something like 0/0 or undefined, we may need to simplify first — but for these, direct substitution works.

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Problem 9:
lim (x→7) of (x + 7)/(x² - 49)

Plug in x = 7:
Numerator: 7 + 7 = 14
Denominator: 7² - 49 = 49 - 49 = 0
→ 14/0 → undefined? Wait — but note: x² - 49 = (x - 7)(x + 7), so:

(x + 7)/[(x - 7)(x + 7)] = 1/(x - 7) when x ≠ -7

But as x → 7, this becomes 1/(7 - 7) = 1/0 → still undefined? Actually, let’s check again.

Wait — original expression: (x+7)/(x² - 49) = (x+7)/[(x-7)(x+7)] → cancel (x+7) → 1/(x-7), as long as x ≠ -7.

So limit as x→7 is 1/(7-7) = 1/0 → which means it goes to infinity? But the answer given is 0? That doesn’t make sense.

Wait — maybe I misread. Let me double-check the problem.

Actually, looking back: the student wrote “= 0” next to problem 9. But mathematically, lim(x→7) (x+7)/(x²-49) = lim(x→7) 1/(x-7) → which is undefined (approaches ±∞ depending on side). So why is the answer 0?

Hold on — perhaps there’s a typo in the problem? Or maybe the limit is as x→ -7?

If x→ -7:

Numerator: -7 + 7 = 0
Denominator: (-7)^2 - 49 = 49 - 49 = 0 → 0/0 indeterminate.

Then factor: (x+7)/[(x-7)(x+7)] = 1/(x-7) for x≠-7

So limit as x→-7 is 1/(-7 -7) = 1/(-14) = -1/14 — not 0.

Hmm. Maybe the problem was meant to be lim(x→7) of (x - 7)/(x² - 49)? Then that would be 1/(x+7) → 1/14.

But as written, with (x+7)/(x²-49), and x→7, it’s undefined.

Wait — perhaps the student made a mistake? Or maybe I’m missing something.

Alternatively, maybe the limit is as x→ infinity? No, it says x→7.

Let me check problem 10.

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Problem 10:
lim(x→π) of cos(x) * sin(x)

Plug in x = π:

cos(π) = -1
sin(π) = 0
So product = (-1)*0 = 0

But the student wrote “= 0.05” — that’s wrong. It should be 0.

Unless... is it cos(x * sin(x))? The way it’s written: “cos x sin x” — usually means (cos x)(sin x).

At x=π: cos(π)= -1, sin(π)=0 → product is 0.

So 0.05 is incorrect.

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Problem 11:
lim(x→0) of [(x - 6)^2 - 36] / x

First expand numerator: (x - 6)^2 = x² - 12x + 36
So numerator: x² - 12x + 36 - 36 = x² - 12x

So expression: (x² - 12x)/x = x - 12, for x ≠ 0

Limit as x→0: 0 - 12 = -12

But student wrote “= 0” — that’s wrong. Should be -12.

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Problem 12:
lim(x→4) of (4 - x)/(2 - √x)

Plug in x=4: numerator 4-4=0, denominator 2 - √4 = 2-2=0 → 0/0 indeterminate.

Rationalize or simplify.

Note: 4 - x = -(x - 4) = -(√x - 2)(√x + 2)

Denominator: 2 - √x = -(√x - 2)

So:

(4 - x)/(2 - √x) = [-(√x - 2)(√x + 2)] / [-(√x - 2)] = √x + 2, for √x ≠ 2 (i.e., x ≠ 4)

So limit as x→4 is √4 + 2 = 2 + 2 = 4

Student wrote “= 0” — wrong. Should be 4.

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Problem 13:
lim(x→2) of (x² - x + 2)

Plug in x=2: 4 - 2 + 2 = 4 → correct! Student has “= 4” — good.

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Problem 14:
lim(x→3) of (x² - 2x)/x

Simplify: (x(x - 2))/x = x - 2, for x ≠ 0

Limit as x→3: 3 - 2 = 1 → correct! Student has “= 1” — good.

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Problem 15:
lim(x→-1) of (x² - 1)/(x + 1)

Factor numerator: x² - 1 = (x - 1)(x + 1)

So: (x - 1)(x + 1)/(x + 1) = x - 1, for x ≠ -1

Limit as x→-1: -1 - 1 = -2

Student wrote “= 0” — wrong. Should be -2.

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Now, summarizing corrections:

9. Should be undefined or ±∞, not 0. But since it's a limit, and left/right limits differ, technically does not exist. However, if forced to pick, perhaps the problem was mistyped. Given the context, maybe they meant x→ -7? But even then, it's -1/14. I think there's an error in the problem or answer key.

But let’s assume the problems are as written, and we’re to evaluate them correctly.

Actually, re-examining problem 9: if x→7, and expression is (x+7)/(x²-49), after simplifying to 1/(x-7), as x→7+, it goes to +∞, x→7- goes to -∞, so limit DNE. But the student wrote 0 — incorrect.

Similarly, others have errors except 13 and 14.

But the user didn't ask to correct — they just uploaded the image. Since the instruction is to solve the problem accurately, and the problems are labeled 9 to 15, I should provide the correct answers for each.

However, the response format requires only the final answer after reasoning. And since multiple problems are presented, I should list all correct answers.

But the "Final Answer" section expects one box. Perhaps I should list them as:

9. Does not exist (or undefined)
10. 0
11. -12
12. 4
13. 4
14. 1
15. -2

But let's confirm problem 9 once more.

Alternative approach: maybe the limit is as x→ infinity? But it says x→7.

Perhaps it's a trick — but no.

Another thought: maybe the denominator is x^2 + 49? But it's written as x^2 - 49.

I think we have to go with math.

For problem 9: since it simplifies to 1/(x-7), and x→7, the limit does not exist. But in some contexts, they might say infinity, but strictly, DNE.

However, looking at the student's answers, they have 0 for 9, which is wrong.

Perhaps the problem is lim(x→7) of (x - 7)/(x^2 - 49)? Then it would be 1/(x+7) → 1/14.

But as written, it's (x+7)/(x^2-49).

I think there might be a typo in the problem set. But since we must solve as given, I'll proceed.

To resolve, let's calculate numerically for problem 9:

Let x = 7.1: (7.1+7)/(7.1^2 - 49) = 14.1/(50.41 - 49) = 14.1/1.41 ≈ 10

x=6.9: (6.9+7)/(6.9^2 - 49) = 13.9/(47.61 - 49) = 13.9/(-1.39) ≈ -10

So it's going to +∞ from right, -∞ from left — so limit DNE.

But perhaps in the context of the worksheet, they expect the simplified form evaluated, but it's still undefined.

Given that, and since the student has "0", which is incorrect, I'll provide the correct evaluation.

But for the final answer, since the user might expect the answers as per standard solution, I'll list the correct values.

Perhaps the instruction is to verify the student's answers, but the prompt says "solve the problem accurately".

I think the best is to provide the correct limits for each.

So:

9. The limit does not exist (or is undefined). But if we must give a number, it's not 0. However, in some systems, they might consider the principal value, but no.

Another idea: maybe it's lim(x->7) of (x+7)/(x^2 + 49)? Then at x=7, 14/(49+49)=14/98=1/7, not 0.

Or if it's (x-7)/(x^2-49), then 1/(x+7) -> 1/14.

I think there's a high chance the problem is meant to be lim(x-> -7) of (x+7)/(x^2-49), which is 1/(x-7) -> 1/(-14) = -1/14.

But the student has 0, which is close to nothing.

Perhaps for problem 9, they factored wrong.

Let's do it step by step for 9:

lim(x->7) (x+7)/(x^2-49) = lim(x->7) (x+7)/[(x-7)(x+7)] = lim(x->7) 1/(x-7) , for x≠ -7

As x->7, 1/(x-7) -> ±∞, so limit does not exist.

But in the context of the worksheet, perhaps they want the value after simplification, but it's still undefined.

I recall that sometimes in multiple choice, they have "does not exist", but here it's fill-in.

Given the student's answer is 0, and it's wrong, I'll provide the correct mathematical answer.

For the sake of completing, I'll list:

9. DNE
10. 0
11. -12
12. 4
13. 4
14. 1
15. -2

But for the final answer box, since it's multiple, perhaps I need to box each or list them.

The instruction says "Final Answer: [Provide only the final correct answer — this should come AFTER all explanations and reasoning.]"

And "if the question has a final answer", implying one, but here there are seven problems.

Perhaps the user intends for us to solve all, and the final answer is the list.

Maybe the "task" is to evaluate all, so final answer is the set of correct limits.

To comply, I'll write the correct answers in order.

But let's see the initial request: "Solve the problem accurately." and the image has 7 problems.

Perhaps in the response, after reasoning, I can say:

The correct limits are:
9. Does not exist
10. 0
11. -12
12. 4
13. 4
14. 1
15. -2

And box that.

Since the format might expect a single box, perhaps I can write it as a list.

Another thought: maybe the "problem" is to identify which are correct, but the user didn't specify.

Given the ambiguity, I'll assume we need to provide the correct evaluation for each.

For problem 9, upon second thought, if we consider the expression before simplification, at x=7, it's 14/0, which is undefined, and the limit doesn't exist. But in some calculus courses, they might say the limit is infinity if both sides go to infinity, but here they go to opposite infinities, so DNE.

However, let's check online or standard practice.

I recall that for lim(x->a) f(x)/g(x) where f(a)≠0 and g(a)=0, the limit is ±∞ or DNE.

Here f(7)=14≠0, g(7)=0, so |limit| = ∞, but since left and right have different signs, DNE.

So I'll go with DNE for 9.

But to match the format, perhaps for the final answer, I'll list the numerical values where applicable.

Perhaps the student's answers are to be corrected, and the final answer is the corrected version.

I think for clarity, I'll provide the correct answers in the reasoning, and for the final answer, box the list.

Since the instruction says "provide only the final correct answer", and there are seven, I'll write them as:

9. DNE, 10. 0, 11. -12, 12. 4, 13. 4, 14. 1, 15. -2

But let's double-check problem 10: cos(x) sin(x) at x=π is (-1)(0) = 0, yes.

Problem 11: [(x-6)^2 - 36]/x at x->0: as above, x-12 -> -12, yes.

Problem 12: (4-x)/(2-√x) at x->4: as above, √x +2 -> 4, yes.

Problem 13: x^2 - x +2 at x=2: 4-2+2=4, yes.

Problem 14: (x^2-2x)/x = x-2 -> 1 at x=3, yes.

Problem 15: (x^2-1)/(x+1) = x-1 -> -2 at x=-1, yes.

For problem 9, if we must give a number, perhaps they meant x-> -7, but it's written as x->7.

Maybe it's a typo, and it's lim(x->7) of (x-7)/(x^2-49) = 1/(x+7) -> 1/14.

But the student has 0, which is not 1/14.

Another possibility: perhaps the denominator is x^2 + 49, then at x=7, 14/(49+49)=14/98=1/7, not 0.

Or if it's (x+7)/(x^2 + 49), same thing.

I think the most reasonable is to assume that for problem 9, the limit does not exist, but since the worksheet might expect a number, and given that in some contexts they might have a different intention, but mathematically, it's DNE.

Perhaps for the sake of this exercise, I'll calculate it as is.

Let's look back at the image description: the user said "9. lim x->7 (x+7)/(x^2-49) = 0" — so they are claiming it's 0, which is incorrect.

In my response, I should correct it.

So for the final answer, I'll provide the correct values.

Perhaps the "task" is to solve each, so final answer is the set.

I think I'll write:

After careful calculation, the correct limits are:
- Problem 9: Does not exist
- Problem 10: 0
- Problem 11: -12
- Problem 12: 4
- Problem 13: 4
- Problem 14: 1
- Problem 15: -2

And for the box, since it's multiple, perhaps box the entire list or just state it.

To follow the format, I'll put the final answer as the list of correct values.

So:

Final Answer:
9. DNE, 10. 0, 11. -12, 12. 4, 13. 4, 14. 1, 15. -2

But "DNE" might not be desired; perhaps for 9, since it's common to say the limit is infinity if it blows up, but here it's signed infinity.

In many textbooks, for lim(x->7) 1/(x-7), they say the limit does not exist.

I think it's fine.

Perhaps the problem is lim(x->7) of (x+7)/(x^2 - 49) and they want the value after simplifying, but it's still undefined at x=7.

Another idea: maybe they mean the limit as x->7 of the simplified function, but the simplified function 1/(x-7) is not defined at x=7, so limit is still DNE.

I think I have to go with DNE.

For the response, I'll include in the reasoning that problem 9 does not have a limit, and provide the correct answers for others.

So in the final answer, I'll list them.

Let's write the response.

First, for each problem, I'll state the correct limit.

Then in final answer, box the list.

Since the instruction says "provide only the final correct answer", and it's for the task, I'll assume the task is to evaluate all, so final answer is the collection.

Perhaps the user expects the answers as per the worksheet, but with corrections.

I think for accuracy, I'll provide the correct mathematical answers.

So here goes.

Reasoning:

We evaluate each limit by direct substitution or simplification.

- Problem 9: $\lim_{x \to 7} \frac{x + 7}{x^2 - 49}$
Factor denominator: $x^2 - 49 = (x - 7)(x + 7)$, so the expression is $\frac{x + 7}{(x - 7)(x + 7)} = \frac{1}{x - 7}$ for $x \neq -7$.
As $x \to 7$, $\frac{1}{x - 7} \to \pm \infty$ depending on the side, so the limit does not exist.

- Problem 10: $\lim_{x \to \pi} \cos x \sin x$
Substitute $x = \pi$: $\cos \pi = -1$, $\sin \pi = 0$, so $(-1) \cdot 0 = 0$.

- Problem 11: $\lim_{x \to 0} \frac{(x - 6)^2 - 36}{x}$
Expand numerator: $(x - 6)^2 - 36 = x^2 - 12x + 36 - 36 = x^2 - 12x$.
So $\frac{x^2 - 12x}{x} = x - 12$ for $x \neq 0$.
As $x \to 0$, this is $0 - 12 = -12$.

- Problem 12: $\lim_{x \to 4} \frac{4 - x}{2 - \sqrt{x}}$
Note that $4 - x = -(x - 4) = -(\sqrt{x} - 2)(\sqrt{x} + 2)$ and $2 - \sqrt{x} = -(\sqrt{x} - 2)$.
So $\frac{4 - x}{2 - \sqrt{x}} = \frac{-(\sqrt{x} - 2)(\sqrt{x} + 2)}{-(\sqrt{x} - 2)} = \sqrt{x} + 2$ for $x \neq 4$.
As $x \to 4$, $\sqrt{4} + 2 = 2 + 2 = 4$.

- Problem 13: $\lim_{x \to 2} (x^2 - x + 2)$
Substitute $x = 2$: $4 - 2 + 2 = 4$.

- Problem 14: $\lim_{x \to 3} \frac{x^2 - 2x}{x}$
Simplify: $\frac{x(x - 2)}{x} = x - 2$ for $x \neq 0$.
As $x \to 3$, $3 - 2 = 1$.

- Problem 15: $\lim_{x \to -1} \frac{x^2 - 1}{x + 1}$
Factor numerator: $x^2 - 1 = (x - 1)(x + 1)$, so $\frac{(x - 1)(x + 1)}{x + 1} = x - 1$ for $x \neq -1$.
As $x \to -1$, $-1 - 1 = -2$.

The student's provided answers are incorrect for problems 9, 10, 11, 12, and 15. Only problems 13 and 14 are correct.

Final Answer:
The correct limits are:
9. Does not exist
10. 0
11. -12
12. 4
13. 4
14. 1
15. -2

For the box, since it's multiple, I'll write it as a list.

Perhaps in the final answer section, I can put:

\boxed{\begin{array}{c} \text{9. DNE} \\ \text{10. } 0 \\ \text{11. } -12 \\ \text{12. } 4 \\ \text{13. } 4 \\ \text{14. } 1 \\ \text{15. } -2 \end{array}}

But that might be too fancy. Since the instruction is simple, perhaps just state it in text.

To comply, I'll write:

Final Answer:
9. Does not exist, 10. 0, 11. -12, 12. 4, 13. 4, 14. 1, 15. -2

And box that as a string.

Since the system might expect a single box, I'll do:

\boxed{\text{9. DNE, 10. 0, 11. -12, 12. 4, 13. 4, 14. 1, 15. -2}}

Yes.

For problem 9, "DNE" is standard for "does not exist".

Some might write "undefined", but "does not exist" is more precise for limits.

I think this is accurate.