How to calculate $lim_{xto 0} frac{sin{(pi sqrt {x+1} )}}{x}$ without L'Hospital's rule? [closed]
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
edited Nov 15 at 19:42
Bernard
118k638112
asked Nov 15 at 19:10
absalon
11
closed as off-topic by TheSimpliFire, Saad, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32
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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
edited Nov 15 at 19:42
Bernard
118k638112
asked Nov 15 at 19:10
absalon
11
closed as off-topic by TheSimpliFire, Saad, 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, Saad, 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 |
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
edited Nov 15 at 19:42
Bernard
118k638112
asked Nov 15 at 19:10
absalon
11
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
118k638112
asked Nov 15 at 19:10
absalon
11
edited Nov 15 at 19:42
Bernard
118k638112
asked Nov 15 at 19:10
absalon
11
edited Nov 15 at 19:42
Bernard
118k638112
edited Nov 15 at 19:42
Bernard
118k638112
edited Nov 15 at 19:42
Bernard
118k638112
118k638112
asked Nov 15 at 19:10
absalon
11
asked Nov 15 at 19:10
absalon
11
asked Nov 15 at 19:10
absalon
11
11
closed as off-topic by TheSimpliFire, Saad, 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, Saad, 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, Saad, 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, Saad, 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 |
add a comment |
2 Answers
2
active
oldest
votes
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
191k28224439
answered Nov 15 at 19:15
Nosrati
26.4k62353
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 |
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
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
answered Nov 15 at 19:59
gimusi
1
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
191k28224439
answered Nov 15 at 19:15
Nosrati
26.4k62353
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 |
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
191k28224439
answered Nov 15 at 19:15
Nosrati
26.4k62353
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 |
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
191k28224439
answered Nov 15 at 19:15
Nosrati
26.4k62353
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
191k28224439
answered Nov 15 at 19:15
Nosrati
26.4k62353
edited Nov 25 at 18:37
amWhy
191k28224439
edited Nov 25 at 18:37
amWhy
191k28224439
edited Nov 25 at 18:37
amWhy
191k28224439
191k28224439
answered Nov 15 at 19:15
Nosrati
26.4k62353
answered Nov 15 at 19:15
Nosrati
26.4k62353
answered Nov 15 at 19:15
Nosrati
26.4k62353
26.4k62353
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 |
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
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
answered Nov 15 at 19:59
gimusi
1
add a comment |
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
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
answered Nov 15 at 19:59
gimusi
1
add a comment |
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
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
answered Nov 15 at 19:59
gimusi
1
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
- $sqrt{x+1}=1+frac12x +o(x)$
$$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$
answered Nov 15 at 19:59
gimusi
1
edited Nov 15 at 20:06
answered Nov 15 at 19:59
gimusi
1
answered Nov 15 at 19:59
gimusi
1
answered Nov 15 at 19:59
gimusi
1
1
add a comment |
add a comment |
