How to calculate $lim_{xto 0} frac{sin{(pi sqrt {x+1} )}}{x}$ without L'Hospital's rule? [closed]
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I got stuck trying to calculate the following limits:
$$
lim_{xto 0-} frac{sin{(pi sqrt {x+1} )}}{x},quad
lim_{xto 0+} frac{sin{(pi sqrt {x+1} )}}{x}
$$
I've tried to approximate the limit using the three functions method, but had no luck. It is also prohibited to use L'Hospital's rule.
calculus limits limits-without-lhopital
closed as off-topic by TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32
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up vote
-3
down vote
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I got stuck trying to calculate the following limits:
$$
lim_{xto 0-} frac{sin{(pi sqrt {x+1} )}}{x},quad
lim_{xto 0+} frac{sin{(pi sqrt {x+1} )}}{x}
$$
I've tried to approximate the limit using the three functions method, but had no luck. It is also prohibited to use L'Hospital's rule.
calculus limits limits-without-lhopital
closed as off-topic by TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
I got stuck trying to calculate the following limits:
$$
lim_{xto 0-} frac{sin{(pi sqrt {x+1} )}}{x},quad
lim_{xto 0+} frac{sin{(pi sqrt {x+1} )}}{x}
$$
I've tried to approximate the limit using the three functions method, but had no luck. It is also prohibited to use L'Hospital's rule.
calculus limits limits-without-lhopital
I got stuck trying to calculate the following limits:
$$
lim_{xto 0-} frac{sin{(pi sqrt {x+1} )}}{x},quad
lim_{xto 0+} frac{sin{(pi sqrt {x+1} )}}{x}
$$
I've tried to approximate the limit using the three functions method, but had no luck. It is also prohibited to use L'Hospital's rule.
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited Nov 15 at 19:42
Bernard
116k637108
116k637108
asked Nov 15 at 19:10
absalon
11
11
closed as off-topic by TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
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2 Answers
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In both cases
$$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$
Why does the second factor converge to pi*(-1/2)?
– absalon
Nov 15 at 19:26
@absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
– hamam_Abdallah
Nov 15 at 19:30
$$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
– Nosrati
Nov 15 at 19:31
When you are displaying a formula or equation, no need for "dfrac" (overkill), as in$$dfrac xy$$. The result can be had via$$frac xy$$
– amWhy
Nov 25 at 18:38
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up vote
0
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As an alternative by $f(x)=sin (pisqrt{x+1})$ we have
$$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$
or by binomial first order expansion
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
In both cases
$$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$
Why does the second factor converge to pi*(-1/2)?
– absalon
Nov 15 at 19:26
@absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
– hamam_Abdallah
Nov 15 at 19:30
$$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
– Nosrati
Nov 15 at 19:31
When you are displaying a formula or equation, no need for "dfrac" (overkill), as in$$dfrac xy$$. The result can be had via$$frac xy$$
– amWhy
Nov 25 at 18:38
add a comment |
up vote
3
down vote
accepted
In both cases
$$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$
Why does the second factor converge to pi*(-1/2)?
– absalon
Nov 15 at 19:26
@absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
– hamam_Abdallah
Nov 15 at 19:30
$$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
– Nosrati
Nov 15 at 19:31
When you are displaying a formula or equation, no need for "dfrac" (overkill), as in$$dfrac xy$$. The result can be had via$$frac xy$$
– amWhy
Nov 25 at 18:38
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In both cases
$$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$
In both cases
$$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$
edited Nov 25 at 18:37
amWhy
191k27223439
191k27223439
answered Nov 15 at 19:15
Nosrati
26.2k62353
26.2k62353
Why does the second factor converge to pi*(-1/2)?
– absalon
Nov 15 at 19:26
@absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
– hamam_Abdallah
Nov 15 at 19:30
$$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
– Nosrati
Nov 15 at 19:31
When you are displaying a formula or equation, no need for "dfrac" (overkill), as in$$dfrac xy$$. The result can be had via$$frac xy$$
– amWhy
Nov 25 at 18:38
add a comment |
Why does the second factor converge to pi*(-1/2)?
– absalon
Nov 15 at 19:26
@absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
– hamam_Abdallah
Nov 15 at 19:30
$$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
– Nosrati
Nov 15 at 19:31
When you are displaying a formula or equation, no need for "dfrac" (overkill), as in$$dfrac xy$$. The result can be had via$$frac xy$$
– amWhy
Nov 25 at 18:38
Why does the second factor converge to pi*(-1/2)?
– absalon
Nov 15 at 19:26
Why does the second factor converge to pi*(-1/2)?
– absalon
Nov 15 at 19:26
@absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
– hamam_Abdallah
Nov 15 at 19:30
@absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
– hamam_Abdallah
Nov 15 at 19:30
$$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
– Nosrati
Nov 15 at 19:31
$$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
– Nosrati
Nov 15 at 19:31
When you are displaying a formula or equation, no need for "dfrac" (overkill), as in
$$dfrac xy$$. The result can be had via $$frac xy$$– amWhy
Nov 25 at 18:38
When you are displaying a formula or equation, no need for "dfrac" (overkill), as in
$$dfrac xy$$. The result can be had via $$frac xy$$– amWhy
Nov 25 at 18:38
add a comment |
up vote
0
down vote
As an alternative by $f(x)=sin (pisqrt{x+1})$ we have
$$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$
or by binomial first order expansion
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
add a comment |
up vote
0
down vote
As an alternative by $f(x)=sin (pisqrt{x+1})$ we have
$$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$
or by binomial first order expansion
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
add a comment |
up vote
0
down vote
up vote
0
down vote
As an alternative by $f(x)=sin (pisqrt{x+1})$ we have
$$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$
or by binomial first order expansion
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
As an alternative by $f(x)=sin (pisqrt{x+1})$ we have
$$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$
or by binomial first order expansion
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
edited Nov 15 at 20:06
answered Nov 15 at 19:59
gimusi
90.6k74495
90.6k74495
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