Tricky limit algebra
up vote
0
down vote
favorite
Could someone please help me figure out the following algbera?
I really don't understand any of the steps.
$$
begin{align*}
lim_{xto 0} frac{1-sqrt{1-x^2}}{x}
&= lim_{xto 0} frac{(1-sqrt{1-x^2})(1+sqrt{1-x^2})}{x(1+sqrt{1-x^2})} \
& = lim_{xto 0} frac{1-(1-x^2)}{x(1+sqrt{1-x^2)}} \
&= lim_{xto 0}frac{x}{1+sqrt{1-x^2}} \
end{align*}
$$
calculus real-analysis limits
edited Nov 13 at 18:49
Stella Biderman
26.4k63175
asked Nov 13 at 18:18
Miguel Cumming-Romo
93
add a comment |
up vote
0
down vote
favorite
Could someone please help me figure out the following algbera?
I really don't understand any of the steps.
$$
begin{align*}
lim_{xto 0} frac{1-sqrt{1-x^2}}{x}
&= lim_{xto 0} frac{(1-sqrt{1-x^2})(1+sqrt{1-x^2})}{x(1+sqrt{1-x^2})} \
& = lim_{xto 0} frac{1-(1-x^2)}{x(1+sqrt{1-x^2)}} \
&= lim_{xto 0}frac{x}{1+sqrt{1-x^2}} \
end{align*}
$$
calculus real-analysis limits
edited Nov 13 at 18:49
Stella Biderman
26.4k63175
asked Nov 13 at 18:18
Miguel Cumming-Romo
93
What about this do you not get? The steps are (in order) multiplying the numerator and denominator by an expression, evaluating the multiplication, simplifying the numerator.
– Rushabh Mehta
Nov 13 at 18:21
I don't get how the square root expressions are being used. For example, in the first step, is what's multiplied to the expression equal to 1? If so, how?
– Miguel Cumming-Romo
Nov 13 at 18:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Could someone please help me figure out the following algbera?
I really don't understand any of the steps.
$$
begin{align*}
lim_{xto 0} frac{1-sqrt{1-x^2}}{x}
&= lim_{xto 0} frac{(1-sqrt{1-x^2})(1+sqrt{1-x^2})}{x(1+sqrt{1-x^2})} \
& = lim_{xto 0} frac{1-(1-x^2)}{x(1+sqrt{1-x^2)}} \
&= lim_{xto 0}frac{x}{1+sqrt{1-x^2}} \
end{align*}
$$
calculus real-analysis limits
edited Nov 13 at 18:49
Stella Biderman
26.4k63175
asked Nov 13 at 18:18
Miguel Cumming-Romo
93
Could someone please help me figure out the following algbera?
I really don't understand any of the steps.
$$
begin{align*}
lim_{xto 0} frac{1-sqrt{1-x^2}}{x}
&= lim_{xto 0} frac{(1-sqrt{1-x^2})(1+sqrt{1-x^2})}{x(1+sqrt{1-x^2})} \
& = lim_{xto 0} frac{1-(1-x^2)}{x(1+sqrt{1-x^2)}} \
&= lim_{xto 0}frac{x}{1+sqrt{1-x^2}} \
end{align*}
$$
calculus real-analysis limits
calculus real-analysis limits
edited Nov 13 at 18:49
Stella Biderman
26.4k63175
asked Nov 13 at 18:18
Miguel Cumming-Romo
93
edited Nov 13 at 18:49
Stella Biderman
26.4k63175
asked Nov 13 at 18:18
Miguel Cumming-Romo
93
edited Nov 13 at 18:49
Stella Biderman
26.4k63175
edited Nov 13 at 18:49
Stella Biderman
26.4k63175
edited Nov 13 at 18:49
Stella Biderman
26.4k63175
26.4k63175
asked Nov 13 at 18:18
Miguel Cumming-Romo
93
asked Nov 13 at 18:18
Miguel Cumming-Romo
93
asked Nov 13 at 18:18
Miguel Cumming-Romo
93
93
What about this do you not get? The steps are (in order) multiplying the numerator and denominator by an expression, evaluating the multiplication, simplifying the numerator.
– Rushabh Mehta
Nov 13 at 18:21
I don't get how the square root expressions are being used. For example, in the first step, is what's multiplied to the expression equal to 1? If so, how?
– Miguel Cumming-Romo
Nov 13 at 18:23
add a comment |
What about this do you not get? The steps are (in order) multiplying the numerator and denominator by an expression, evaluating the multiplication, simplifying the numerator.
– Rushabh Mehta
Nov 13 at 18:21
I don't get how the square root expressions are being used. For example, in the first step, is what's multiplied to the expression equal to 1? If so, how?
– Miguel Cumming-Romo
Nov 13 at 18:23
What about this do you not get? The steps are (in order) multiplying the numerator and denominator by an expression, evaluating the multiplication, simplifying the numerator.
– Rushabh Mehta
Nov 13 at 18:21
What about this do you not get? The steps are (in order) multiplying the numerator and denominator by an expression, evaluating the multiplication, simplifying the numerator.
– Rushabh Mehta
Nov 13 at 18:21
I don't get how the square root expressions are being used. For example, in the first step, is what's multiplied to the expression equal to 1? If so, how?
– Miguel Cumming-Romo
Nov 13 at 18:23
I don't get how the square root expressions are being used. For example, in the first step, is what's multiplied to the expression equal to 1? If so, how?
– Miguel Cumming-Romo
Nov 13 at 18:23
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
If you try to evaluate the limit in its initial form you end up with $$lim_{xto 0} frac{1-sqrt{1-x^2}}{x}=frac{0}{0}$$
which is an indeterminate form.
To get rid of this indetermination, you simply need to multiply the numerator and the denominator by the conjugate of the numerator, which is $1+sqrt{1-x^2}$.
After making some simplifications, the $x$ which caused the denominator's tending to $0$ as $xto0$ causing the indetermination gets deleted. Now the limit becomes:
$$lim_{xto 0}frac{x}{1+sqrt{1-x^2}}=frac{0}{2}=0$$
answered Nov 13 at 18:34
Jevaut
5049
I understand why we do, it's just how that I don't get. For example, why is it that we use conjugate at the numerator but a different expression at the denominator? Are you telling me those two equal to one?
– Miguel Cumming-Romo
Nov 13 at 18:37
If you multiply the numerator and the denominator of a fraction by the same number or expression $kleq 0$, the fraction does not change because $frac{k}{k}=1$. Also, the multiplying by the conjugate is a standard step for limits which involve expressions with roots. Roots go away after applying the identity $(a-b)(a+b)=a^2-b^2$ and usually the terms which zero out the denominator go away too. After all those steps, the limit is easily computable.
– Jevaut
Nov 13 at 20:29
add a comment |
up vote
0
down vote
The steps you have given aren't really to do with the limits at all, it's just a question about manipulating the algebraic expression.
The first step essentially is "multiplying by one", probably our most useful tool in manipulating fractions. It is not hard to see that $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}=1$, and clearly anything multiplied by one remains the same (due to 1 being the identity element for the binary operation of scalar multiplication). But why $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}$?
The term $1+sqrt{1-x^2}$ is said to be the "conjugate" of $1-sqrt{1-x^2}$, and their product is said to be the difference of two squares (i.e. $(1+sqrt{1-x^2})(1-sqrt{1-x^2})=(1^2-(sqrt{1-x^2})^2=(1-(1-x^2))$ - see https://en.wikipedia.org/wiki/Difference_of_two_squares for more information on this.
So we have chosen this product to engineer a numerator without a square root in it.
It is clear that multiplying out the numerator gives us our desired form $$frac{(1+sqrt{1-x^2})(1-sqrt{1-x^2})}{x(1+sqrt{x^2})} = frac{1-(1-x^2)}{x(1+sqrt{1-x^2})}$$
Now, we can simplify the numerator $1-(1-x^2)=-(-x^2)=x^2$, this gives us the fraction $$frac{x}{1+sqrt{1-x^2}}$$ as required.
answered Nov 13 at 18:37
Joel Biffin
515
Haha, I see now. Wow, how did I not see this. Thanks for your answer!
– Miguel Cumming-Romo
Nov 13 at 18:40
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you try to evaluate the limit in its initial form you end up with $$lim_{xto 0} frac{1-sqrt{1-x^2}}{x}=frac{0}{0}$$
which is an indeterminate form.
To get rid of this indetermination, you simply need to multiply the numerator and the denominator by the conjugate of the numerator, which is $1+sqrt{1-x^2}$.
After making some simplifications, the $x$ which caused the denominator's tending to $0$ as $xto0$ causing the indetermination gets deleted. Now the limit becomes:
$$lim_{xto 0}frac{x}{1+sqrt{1-x^2}}=frac{0}{2}=0$$
answered Nov 13 at 18:34
Jevaut
5049
I understand why we do, it's just how that I don't get. For example, why is it that we use conjugate at the numerator but a different expression at the denominator? Are you telling me those two equal to one?
– Miguel Cumming-Romo
Nov 13 at 18:37
If you multiply the numerator and the denominator of a fraction by the same number or expression $kleq 0$, the fraction does not change because $frac{k}{k}=1$. Also, the multiplying by the conjugate is a standard step for limits which involve expressions with roots. Roots go away after applying the identity $(a-b)(a+b)=a^2-b^2$ and usually the terms which zero out the denominator go away too. After all those steps, the limit is easily computable.
– Jevaut
Nov 13 at 20:29
add a comment |
up vote
0
down vote
If you try to evaluate the limit in its initial form you end up with $$lim_{xto 0} frac{1-sqrt{1-x^2}}{x}=frac{0}{0}$$
which is an indeterminate form.
To get rid of this indetermination, you simply need to multiply the numerator and the denominator by the conjugate of the numerator, which is $1+sqrt{1-x^2}$.
After making some simplifications, the $x$ which caused the denominator's tending to $0$ as $xto0$ causing the indetermination gets deleted. Now the limit becomes:
$$lim_{xto 0}frac{x}{1+sqrt{1-x^2}}=frac{0}{2}=0$$
answered Nov 13 at 18:34
Jevaut
5049
I understand why we do, it's just how that I don't get. For example, why is it that we use conjugate at the numerator but a different expression at the denominator? Are you telling me those two equal to one?
– Miguel Cumming-Romo
Nov 13 at 18:37
If you multiply the numerator and the denominator of a fraction by the same number or expression $kleq 0$, the fraction does not change because $frac{k}{k}=1$. Also, the multiplying by the conjugate is a standard step for limits which involve expressions with roots. Roots go away after applying the identity $(a-b)(a+b)=a^2-b^2$ and usually the terms which zero out the denominator go away too. After all those steps, the limit is easily computable.
– Jevaut
Nov 13 at 20:29
add a comment |
up vote
0
down vote
up vote
0
down vote
If you try to evaluate the limit in its initial form you end up with $$lim_{xto 0} frac{1-sqrt{1-x^2}}{x}=frac{0}{0}$$
which is an indeterminate form.
To get rid of this indetermination, you simply need to multiply the numerator and the denominator by the conjugate of the numerator, which is $1+sqrt{1-x^2}$.
After making some simplifications, the $x$ which caused the denominator's tending to $0$ as $xto0$ causing the indetermination gets deleted. Now the limit becomes:
$$lim_{xto 0}frac{x}{1+sqrt{1-x^2}}=frac{0}{2}=0$$
answered Nov 13 at 18:34
Jevaut
5049
If you try to evaluate the limit in its initial form you end up with $$lim_{xto 0} frac{1-sqrt{1-x^2}}{x}=frac{0}{0}$$
which is an indeterminate form.
To get rid of this indetermination, you simply need to multiply the numerator and the denominator by the conjugate of the numerator, which is $1+sqrt{1-x^2}$.
After making some simplifications, the $x$ which caused the denominator's tending to $0$ as $xto0$ causing the indetermination gets deleted. Now the limit becomes:
$$lim_{xto 0}frac{x}{1+sqrt{1-x^2}}=frac{0}{2}=0$$
answered Nov 13 at 18:34
Jevaut
5049
answered Nov 13 at 18:34
Jevaut
5049
answered Nov 13 at 18:34
Jevaut
5049
answered Nov 13 at 18:34
Jevaut
5049
5049
I understand why we do, it's just how that I don't get. For example, why is it that we use conjugate at the numerator but a different expression at the denominator? Are you telling me those two equal to one?
– Miguel Cumming-Romo
Nov 13 at 18:37
If you multiply the numerator and the denominator of a fraction by the same number or expression $kleq 0$, the fraction does not change because $frac{k}{k}=1$. Also, the multiplying by the conjugate is a standard step for limits which involve expressions with roots. Roots go away after applying the identity $(a-b)(a+b)=a^2-b^2$ and usually the terms which zero out the denominator go away too. After all those steps, the limit is easily computable.
– Jevaut
Nov 13 at 20:29
add a comment |
I understand why we do, it's just how that I don't get. For example, why is it that we use conjugate at the numerator but a different expression at the denominator? Are you telling me those two equal to one?
– Miguel Cumming-Romo
Nov 13 at 18:37
If you multiply the numerator and the denominator of a fraction by the same number or expression $kleq 0$, the fraction does not change because $frac{k}{k}=1$. Also, the multiplying by the conjugate is a standard step for limits which involve expressions with roots. Roots go away after applying the identity $(a-b)(a+b)=a^2-b^2$ and usually the terms which zero out the denominator go away too. After all those steps, the limit is easily computable.
– Jevaut
Nov 13 at 20:29
I understand why we do, it's just how that I don't get. For example, why is it that we use conjugate at the numerator but a different expression at the denominator? Are you telling me those two equal to one?
– Miguel Cumming-Romo
Nov 13 at 18:37
I understand why we do, it's just how that I don't get. For example, why is it that we use conjugate at the numerator but a different expression at the denominator? Are you telling me those two equal to one?
– Miguel Cumming-Romo
Nov 13 at 18:37
If you multiply the numerator and the denominator of a fraction by the same number or expression $kleq 0$, the fraction does not change because $frac{k}{k}=1$. Also, the multiplying by the conjugate is a standard step for limits which involve expressions with roots. Roots go away after applying the identity $(a-b)(a+b)=a^2-b^2$ and usually the terms which zero out the denominator go away too. After all those steps, the limit is easily computable.
– Jevaut
Nov 13 at 20:29
If you multiply the numerator and the denominator of a fraction by the same number or expression $kleq 0$, the fraction does not change because $frac{k}{k}=1$. Also, the multiplying by the conjugate is a standard step for limits which involve expressions with roots. Roots go away after applying the identity $(a-b)(a+b)=a^2-b^2$ and usually the terms which zero out the denominator go away too. After all those steps, the limit is easily computable.
– Jevaut
Nov 13 at 20:29
add a comment |
up vote
0
down vote
The steps you have given aren't really to do with the limits at all, it's just a question about manipulating the algebraic expression.
The first step essentially is "multiplying by one", probably our most useful tool in manipulating fractions. It is not hard to see that $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}=1$, and clearly anything multiplied by one remains the same (due to 1 being the identity element for the binary operation of scalar multiplication). But why $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}$?
The term $1+sqrt{1-x^2}$ is said to be the "conjugate" of $1-sqrt{1-x^2}$, and their product is said to be the difference of two squares (i.e. $(1+sqrt{1-x^2})(1-sqrt{1-x^2})=(1^2-(sqrt{1-x^2})^2=(1-(1-x^2))$ - see https://en.wikipedia.org/wiki/Difference_of_two_squares for more information on this.
So we have chosen this product to engineer a numerator without a square root in it.
It is clear that multiplying out the numerator gives us our desired form $$frac{(1+sqrt{1-x^2})(1-sqrt{1-x^2})}{x(1+sqrt{x^2})} = frac{1-(1-x^2)}{x(1+sqrt{1-x^2})}$$
Now, we can simplify the numerator $1-(1-x^2)=-(-x^2)=x^2$, this gives us the fraction $$frac{x}{1+sqrt{1-x^2}}$$ as required.
answered Nov 13 at 18:37
Joel Biffin
515
Haha, I see now. Wow, how did I not see this. Thanks for your answer!
– Miguel Cumming-Romo
Nov 13 at 18:40
add a comment |
up vote
0
down vote
The steps you have given aren't really to do with the limits at all, it's just a question about manipulating the algebraic expression.
The first step essentially is "multiplying by one", probably our most useful tool in manipulating fractions. It is not hard to see that $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}=1$, and clearly anything multiplied by one remains the same (due to 1 being the identity element for the binary operation of scalar multiplication). But why $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}$?
The term $1+sqrt{1-x^2}$ is said to be the "conjugate" of $1-sqrt{1-x^2}$, and their product is said to be the difference of two squares (i.e. $(1+sqrt{1-x^2})(1-sqrt{1-x^2})=(1^2-(sqrt{1-x^2})^2=(1-(1-x^2))$ - see https://en.wikipedia.org/wiki/Difference_of_two_squares for more information on this.
So we have chosen this product to engineer a numerator without a square root in it.
It is clear that multiplying out the numerator gives us our desired form $$frac{(1+sqrt{1-x^2})(1-sqrt{1-x^2})}{x(1+sqrt{x^2})} = frac{1-(1-x^2)}{x(1+sqrt{1-x^2})}$$
Now, we can simplify the numerator $1-(1-x^2)=-(-x^2)=x^2$, this gives us the fraction $$frac{x}{1+sqrt{1-x^2}}$$ as required.
answered Nov 13 at 18:37
Joel Biffin
515
Haha, I see now. Wow, how did I not see this. Thanks for your answer!
– Miguel Cumming-Romo
Nov 13 at 18:40
add a comment |
up vote
0
down vote
up vote
0
down vote
The steps you have given aren't really to do with the limits at all, it's just a question about manipulating the algebraic expression.
The first step essentially is "multiplying by one", probably our most useful tool in manipulating fractions. It is not hard to see that $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}=1$, and clearly anything multiplied by one remains the same (due to 1 being the identity element for the binary operation of scalar multiplication). But why $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}$?
The term $1+sqrt{1-x^2}$ is said to be the "conjugate" of $1-sqrt{1-x^2}$, and their product is said to be the difference of two squares (i.e. $(1+sqrt{1-x^2})(1-sqrt{1-x^2})=(1^2-(sqrt{1-x^2})^2=(1-(1-x^2))$ - see https://en.wikipedia.org/wiki/Difference_of_two_squares for more information on this.
So we have chosen this product to engineer a numerator without a square root in it.
It is clear that multiplying out the numerator gives us our desired form $$frac{(1+sqrt{1-x^2})(1-sqrt{1-x^2})}{x(1+sqrt{x^2})} = frac{1-(1-x^2)}{x(1+sqrt{1-x^2})}$$
Now, we can simplify the numerator $1-(1-x^2)=-(-x^2)=x^2$, this gives us the fraction $$frac{x}{1+sqrt{1-x^2}}$$ as required.
answered Nov 13 at 18:37
Joel Biffin
515
The steps you have given aren't really to do with the limits at all, it's just a question about manipulating the algebraic expression.
The first step essentially is "multiplying by one", probably our most useful tool in manipulating fractions. It is not hard to see that $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}=1$, and clearly anything multiplied by one remains the same (due to 1 being the identity element for the binary operation of scalar multiplication). But why $frac{1+sqrt{1-x^2}}{1+sqrt{1-x^2}}$?
The term $1+sqrt{1-x^2}$ is said to be the "conjugate" of $1-sqrt{1-x^2}$, and their product is said to be the difference of two squares (i.e. $(1+sqrt{1-x^2})(1-sqrt{1-x^2})=(1^2-(sqrt{1-x^2})^2=(1-(1-x^2))$ - see https://en.wikipedia.org/wiki/Difference_of_two_squares for more information on this.
So we have chosen this product to engineer a numerator without a square root in it.
It is clear that multiplying out the numerator gives us our desired form $$frac{(1+sqrt{1-x^2})(1-sqrt{1-x^2})}{x(1+sqrt{x^2})} = frac{1-(1-x^2)}{x(1+sqrt{1-x^2})}$$
Now, we can simplify the numerator $1-(1-x^2)=-(-x^2)=x^2$, this gives us the fraction $$frac{x}{1+sqrt{1-x^2}}$$ as required.
answered Nov 13 at 18:37
Joel Biffin
515
answered Nov 13 at 18:37
Joel Biffin
515
answered Nov 13 at 18:37
Joel Biffin
515
answered Nov 13 at 18:37
Joel Biffin
515
515
Haha, I see now. Wow, how did I not see this. Thanks for your answer!
– Miguel Cumming-Romo
Nov 13 at 18:40
add a comment |
Haha, I see now. Wow, how did I not see this. Thanks for your answer!
– Miguel Cumming-Romo
Nov 13 at 18:40
Haha, I see now. Wow, how did I not see this. Thanks for your answer!
– Miguel Cumming-Romo
Nov 13 at 18:40
Haha, I see now. Wow, how did I not see this. Thanks for your answer!
– Miguel Cumming-Romo
Nov 13 at 18:40
add a comment |
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What about this do you not get? The steps are (in order) multiplying the numerator and denominator by an expression, evaluating the multiplication, simplifying the numerator.
– Rushabh Mehta
Nov 13 at 18:21
I don't get how the square root expressions are being used. For example, in the first step, is what's multiplied to the expression equal to 1? If so, how?
– Miguel Cumming-Romo
Nov 13 at 18:23