Proof using derivative information to find limitlimit $lim_x to 49 fracsqrtx-7x-49 .$Can you help me calculate a limit?Trying to find limit by rationalizing numerator with square rootLimit of a functionLimit of real logarithmFind $lim_x rightarrow 1 fracsqrt2x^2-1-1x^3-1$ without using L'Hôpital's ruleHow to calculate this limit $lim_xtoinftyfracsqrt9x^2+2x-3(8x^5-6x+1)^frac 13$Integral proof using comparison theorem
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Proof using derivative information to find limit
limit $lim_x to 49 fracsqrtx-7x-49 .$Can you help me calculate a limit?Trying to find limit by rationalizing numerator with square rootLimit of a functionLimit of real logarithmFind $lim_x rightarrow 1 fracsqrt2x^2-1-1x^3-1$ without using L'Hôpital's ruleHow to calculate this limit $lim_xtoinftyfracsqrt9x^2+2x-3(8x^5-6x+1)^frac 13$Integral proof using comparison theorem
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$begingroup$
This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.
The exercise is:
If $f'(a)=1$ for $a>0$, find $lim_x to a fracf(x)-f(a)sqrtx-sqrta$.
What came to my mind was to rationalize the denominator.
$$lim_x to a fracf(x)-f(a)sqrtx-sqrta$$
$$=lim_x to a fracf(x)-f(a)sqrtx-sqrtacdot fracsqrtx+sqrtasqrtx+sqrta$$
$$=lim_x to a frac(f(x)-f(a))(sqrtx+sqrta)x-a$$
$$=lim_x to a left(fracf(x)-f(a)x-acdot (sqrtx+sqrta)right)$$
$$=lim_x to a fracf(x)-f(a)x-acdot lim_x to a(sqrtx+sqrta)$$
$$=f'(a)cdot lim_x to a(sqrtx+sqrta)$$
$$=1cdot lim_x to a(sqrtx+sqrta)$$
$$=lim_x to a(sqrtx+sqrta)$$
$$=sqrta+sqrta$$
$$=2sqrta$$
calculus limits derivatives
asked 8 hours ago
gi2302gi2302
1207 bronze badges
$endgroup$
add a comment
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$begingroup$
This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.
The exercise is:
If $f'(a)=1$ for $a>0$, find $lim_x to a fracf(x)-f(a)sqrtx-sqrta$.
What came to my mind was to rationalize the denominator.
$$lim_x to a fracf(x)-f(a)sqrtx-sqrta$$
$$=lim_x to a fracf(x)-f(a)sqrtx-sqrtacdot fracsqrtx+sqrtasqrtx+sqrta$$
$$=lim_x to a frac(f(x)-f(a))(sqrtx+sqrta)x-a$$
$$=lim_x to a left(fracf(x)-f(a)x-acdot (sqrtx+sqrta)right)$$
$$=lim_x to a fracf(x)-f(a)x-acdot lim_x to a(sqrtx+sqrta)$$
$$=f'(a)cdot lim_x to a(sqrtx+sqrta)$$
$$=1cdot lim_x to a(sqrtx+sqrta)$$
$$=lim_x to a(sqrtx+sqrta)$$
$$=sqrta+sqrta$$
$$=2sqrta$$
calculus limits derivatives
asked 8 hours ago
gi2302gi2302
1207 bronze badges
$endgroup$
add a comment
|
$begingroup$
This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.
The exercise is:
If $f'(a)=1$ for $a>0$, find $lim_x to a fracf(x)-f(a)sqrtx-sqrta$.
What came to my mind was to rationalize the denominator.
$$lim_x to a fracf(x)-f(a)sqrtx-sqrta$$
$$=lim_x to a fracf(x)-f(a)sqrtx-sqrtacdot fracsqrtx+sqrtasqrtx+sqrta$$
$$=lim_x to a frac(f(x)-f(a))(sqrtx+sqrta)x-a$$
$$=lim_x to a left(fracf(x)-f(a)x-acdot (sqrtx+sqrta)right)$$
$$=lim_x to a fracf(x)-f(a)x-acdot lim_x to a(sqrtx+sqrta)$$
$$=f'(a)cdot lim_x to a(sqrtx+sqrta)$$
$$=1cdot lim_x to a(sqrtx+sqrta)$$
$$=lim_x to a(sqrtx+sqrta)$$
$$=sqrta+sqrta$$
$$=2sqrta$$
calculus limits derivatives
asked 8 hours ago
gi2302gi2302
1207 bronze badges
$endgroup$
This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.
The exercise is:
If $f'(a)=1$ for $a>0$, find $lim_x to a fracf(x)-f(a)sqrtx-sqrta$.
What came to my mind was to rationalize the denominator.
$$lim_x to a fracf(x)-f(a)sqrtx-sqrta$$
$$=lim_x to a fracf(x)-f(a)sqrtx-sqrtacdot fracsqrtx+sqrtasqrtx+sqrta$$
$$=lim_x to a frac(f(x)-f(a))(sqrtx+sqrta)x-a$$
$$=lim_x to a left(fracf(x)-f(a)x-acdot (sqrtx+sqrta)right)$$
$$=lim_x to a fracf(x)-f(a)x-acdot lim_x to a(sqrtx+sqrta)$$
$$=f'(a)cdot lim_x to a(sqrtx+sqrta)$$
$$=1cdot lim_x to a(sqrtx+sqrta)$$
$$=lim_x to a(sqrtx+sqrta)$$
$$=sqrta+sqrta$$
$$=2sqrta$$
calculus limits derivatives
calculus limits derivatives
asked 8 hours ago
gi2302gi2302
1207 bronze badges
asked 8 hours ago
gi2302gi2302
1207 bronze badges
edited 8 hours ago
gi2302
asked 8 hours ago
gi2302gi2302
1207 bronze badges
asked 8 hours ago
gi2302gi2302
1207 bronze badges
asked 8 hours ago
gi2302gi2302
1207 bronze badges
1207 bronze badges
add a comment
|
add a comment
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1 Answer
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$begingroup$
You made a mistake. (It's not a big one but it impacts your answer).
On Line $3$, you should have had $sqrtx+sqrta$ in numerator.
Now, redo the steps (easy), and you'll end up with $2sqrta$.
answered 8 hours ago
Saketh MalyalaSaketh Malyala
11.6k17 silver badges39 bronze badges
$endgroup$
$begingroup$
great.. i will fix it.. thanks so much!
$endgroup$
– gi2302
8 hours ago
add a comment
|
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
You made a mistake. (It's not a big one but it impacts your answer).
On Line $3$, you should have had $sqrtx+sqrta$ in numerator.
Now, redo the steps (easy), and you'll end up with $2sqrta$.
answered 8 hours ago
Saketh MalyalaSaketh Malyala
11.6k17 silver badges39 bronze badges
$endgroup$
$begingroup$
great.. i will fix it.. thanks so much!
$endgroup$
– gi2302
8 hours ago
add a comment
|
$begingroup$
You made a mistake. (It's not a big one but it impacts your answer).
On Line $3$, you should have had $sqrtx+sqrta$ in numerator.
Now, redo the steps (easy), and you'll end up with $2sqrta$.
answered 8 hours ago
Saketh MalyalaSaketh Malyala
11.6k17 silver badges39 bronze badges
$endgroup$
$begingroup$
great.. i will fix it.. thanks so much!
$endgroup$
– gi2302
8 hours ago
add a comment
|
$begingroup$
You made a mistake. (It's not a big one but it impacts your answer).
On Line $3$, you should have had $sqrtx+sqrta$ in numerator.
Now, redo the steps (easy), and you'll end up with $2sqrta$.
answered 8 hours ago
Saketh MalyalaSaketh Malyala
11.6k17 silver badges39 bronze badges
$endgroup$
You made a mistake. (It's not a big one but it impacts your answer).
On Line $3$, you should have had $sqrtx+sqrta$ in numerator.
Now, redo the steps (easy), and you'll end up with $2sqrta$.
answered 8 hours ago
Saketh MalyalaSaketh Malyala
11.6k17 silver badges39 bronze badges
answered 8 hours ago
Saketh MalyalaSaketh Malyala
11.6k17 silver badges39 bronze badges
answered 8 hours ago
Saketh MalyalaSaketh Malyala
11.6k17 silver badges39 bronze badges
answered 8 hours ago
Saketh MalyalaSaketh Malyala
11.6k17 silver badges39 bronze badges
11.6k17 silver badges39 bronze badges
$begingroup$
great.. i will fix it.. thanks so much!
$endgroup$
– gi2302
8 hours ago
add a comment
|
$begingroup$
great.. i will fix it.. thanks so much!
$endgroup$
– gi2302
8 hours ago
$begingroup$
great.. i will fix it.. thanks so much!
$endgroup$
– gi2302
8 hours ago
$begingroup$
great.. i will fix it.. thanks so much!
$endgroup$
– gi2302
8 hours ago
add a comment
|
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