How to calculate $lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$$
What I tried to do was to divide every term my $x^2$:
$$frac{frac{1}{x} + frac{2}{x^2}}{sqrt{9+frac{1}{x^2}}}$$
Then I calculated the limits of the numerator and denominator separately, which gave:
$$frac{0}{3}$$
For some reason though, it appears that the right answer is $frac{1}{3}$? Can someone explain me what's wrong with my solution?
limits
asked Dec 2 '18 at 23:59
Trey
304113
add a comment |
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$$
What I tried to do was to divide every term my $x^2$:
$$frac{frac{1}{x} + frac{2}{x^2}}{sqrt{9+frac{1}{x^2}}}$$
Then I calculated the limits of the numerator and denominator separately, which gave:
$$frac{0}{3}$$
For some reason though, it appears that the right answer is $frac{1}{3}$? Can someone explain me what's wrong with my solution?
limits
asked Dec 2 '18 at 23:59
Trey
304113
You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
– T. Bongers
Dec 3 '18 at 0:00
@T.Bongers not really, care to explain?
– Trey
Dec 3 '18 at 0:01
What is $sqrt{x^2}$?
– T. Bongers
Dec 3 '18 at 0:03
(1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
– GEdgar
Dec 3 '18 at 0:11
add a comment |
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$$
What I tried to do was to divide every term my $x^2$:
$$frac{frac{1}{x} + frac{2}{x^2}}{sqrt{9+frac{1}{x^2}}}$$
Then I calculated the limits of the numerator and denominator separately, which gave:
$$frac{0}{3}$$
For some reason though, it appears that the right answer is $frac{1}{3}$? Can someone explain me what's wrong with my solution?
limits
asked Dec 2 '18 at 23:59
Trey
304113
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$$
What I tried to do was to divide every term my $x^2$:
$$frac{frac{1}{x} + frac{2}{x^2}}{sqrt{9+frac{1}{x^2}}}$$
Then I calculated the limits of the numerator and denominator separately, which gave:
$$frac{0}{3}$$
For some reason though, it appears that the right answer is $frac{1}{3}$? Can someone explain me what's wrong with my solution?
limits
limits
asked Dec 2 '18 at 23:59
Trey
304113
asked Dec 2 '18 at 23:59
Trey
304113
asked Dec 2 '18 at 23:59
Trey
304113
asked Dec 2 '18 at 23:59
Trey
304113
asked Dec 2 '18 at 23:59
Trey
304113
304113
You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
– T. Bongers
Dec 3 '18 at 0:00
@T.Bongers not really, care to explain?
– Trey
Dec 3 '18 at 0:01
What is $sqrt{x^2}$?
– T. Bongers
Dec 3 '18 at 0:03
(1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
– GEdgar
Dec 3 '18 at 0:11
add a comment |
You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
– T. Bongers
Dec 3 '18 at 0:00
@T.Bongers not really, care to explain?
– Trey
Dec 3 '18 at 0:01
What is $sqrt{x^2}$?
– T. Bongers
Dec 3 '18 at 0:03
(1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
– GEdgar
Dec 3 '18 at 0:11
You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
– T. Bongers
Dec 3 '18 at 0:00
You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
– T. Bongers
Dec 3 '18 at 0:00
@T.Bongers not really, care to explain?
– Trey
Dec 3 '18 at 0:01
@T.Bongers not really, care to explain?
– Trey
Dec 3 '18 at 0:01
What is $sqrt{x^2}$?
– T. Bongers
Dec 3 '18 at 0:03
What is $sqrt{x^2}$?
– T. Bongers
Dec 3 '18 at 0:03
(1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
– GEdgar
Dec 3 '18 at 0:11
(1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
– GEdgar
Dec 3 '18 at 0:11
add a comment |
2 Answers
2
active
oldest
votes
Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.
However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:
$$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$
answered Dec 3 '18 at 0:07
Ben W
1,770514
Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
– T. Bongers
Dec 3 '18 at 0:08
That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
– gimusi
Dec 3 '18 at 0:09
That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
– Ben W
Dec 3 '18 at 0:11
I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
– gimusi
Dec 3 '18 at 0:27
What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
– Ben W
Dec 3 '18 at 0:34
add a comment |
HINT
We have
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$
answered Dec 3 '18 at 0:06
gimusi
1
add a comment |
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2 Answers
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2 Answers
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Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.
However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:
$$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$
answered Dec 3 '18 at 0:07
Ben W
1,770514
Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
– T. Bongers
Dec 3 '18 at 0:08
That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
– gimusi
Dec 3 '18 at 0:09
That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
– Ben W
Dec 3 '18 at 0:11
I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
– gimusi
Dec 3 '18 at 0:27
What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
– Ben W
Dec 3 '18 at 0:34
add a comment |
Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.
However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:
$$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$
answered Dec 3 '18 at 0:07
Ben W
1,770514
Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
– T. Bongers
Dec 3 '18 at 0:08
That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
– gimusi
Dec 3 '18 at 0:09
That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
– Ben W
Dec 3 '18 at 0:11
I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
– gimusi
Dec 3 '18 at 0:27
What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
– Ben W
Dec 3 '18 at 0:34
add a comment |
Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.
However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:
$$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$
answered Dec 3 '18 at 0:07
Ben W
1,770514
Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.
However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:
$$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$
answered Dec 3 '18 at 0:07
Ben W
1,770514
edited Dec 3 '18 at 0:14
answered Dec 3 '18 at 0:07
Ben W
1,770514
answered Dec 3 '18 at 0:07
Ben W
1,770514
answered Dec 3 '18 at 0:07
Ben W
1,770514
1,770514
Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
– T. Bongers
Dec 3 '18 at 0:08
That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
– gimusi
Dec 3 '18 at 0:09
That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
– Ben W
Dec 3 '18 at 0:11
I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
– gimusi
Dec 3 '18 at 0:27
What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
– Ben W
Dec 3 '18 at 0:34
add a comment |
Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
– T. Bongers
Dec 3 '18 at 0:08
That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
– gimusi
Dec 3 '18 at 0:09
That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
– Ben W
Dec 3 '18 at 0:11
I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
– gimusi
Dec 3 '18 at 0:27
What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
– Ben W
Dec 3 '18 at 0:34
Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
– T. Bongers
Dec 3 '18 at 0:08
Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
– T. Bongers
Dec 3 '18 at 0:08
That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
– gimusi
Dec 3 '18 at 0:09
That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
– gimusi
Dec 3 '18 at 0:09
That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
– Ben W
Dec 3 '18 at 0:11
That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
– Ben W
Dec 3 '18 at 0:11
I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
– gimusi
Dec 3 '18 at 0:27
I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
– gimusi
Dec 3 '18 at 0:27
What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
– Ben W
Dec 3 '18 at 0:34
What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
– Ben W
Dec 3 '18 at 0:34
add a comment |
HINT
We have
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$
answered Dec 3 '18 at 0:06
gimusi
1
add a comment |
HINT
We have
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$
answered Dec 3 '18 at 0:06
gimusi
1
add a comment |
HINT
We have
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$
answered Dec 3 '18 at 0:06
gimusi
1
HINT
We have
$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$
answered Dec 3 '18 at 0:06
gimusi
1
answered Dec 3 '18 at 0:06
gimusi
1
answered Dec 3 '18 at 0:06
gimusi
1
answered Dec 3 '18 at 0:06
gimusi
1
1
add a comment |
add a comment |
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You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
– T. Bongers
Dec 3 '18 at 0:00
@T.Bongers not really, care to explain?
– Trey
Dec 3 '18 at 0:01
What is $sqrt{x^2}$?
– T. Bongers
Dec 3 '18 at 0:03
(1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
– GEdgar
Dec 3 '18 at 0:11