qanda-logo
search-icon
root-icon-orangeSymbol
apple-logoApp Storegoogle-play-logoGoogle Play

Calculator search results

Calculate the limiting value
Answer
circle-check-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
$2$
Using a multiplication expression
$\color{#FF6800}{ \lim\limits_{x\to + ∞} } \left( \color{#FF6800}{\dfrac {2x ^ {2} - 2x - 3} {x ^ {2} - 1}} \right)$
$ $ Find the limit value of the numerator and denominator separately $ $
$\color{#FF6800}{ \lim\limits_{x\to + ∞} } \left( \color{#FF6800}{2x ^ {2} - 2x - 3} \right)\\\color{#FF6800}{ \lim\limits_{x\to + ∞} } \left( \color{#FF6800}{x ^ {2} - 1} \right)$
$\color{#FF6800}{ \lim\limits_{x\to + ∞} } \left( \color{#FF6800}{2x ^ {2} - 2x - 3} \right)\\\lim\limits_{x\to + ∞} \left( x ^ {2} - 1 \right)$
$ $ Calculate the limiting value $ $
$\color{#FF6800}{+ ∞}\\\lim\limits_{x\to + ∞} \left( x ^ {2} - 1 \right)$
$+ ∞\\\color{#FF6800}{ \lim\limits_{x\to + ∞} } \left( \color{#FF6800}{x ^ {2} - 1} \right)$
$ $ Calculate the limiting value $ $
$+ ∞\\\color{#FF6800}{+ ∞}$
$\color{#FF6800}{+ ∞}\\\color{#FF6800}{+ ∞}$
$ $ Since the limit converging is indedterninate form, convert the formula $ $
$\color{#FF6800}{ \lim\limits_{x\to + ∞} } \left( \color{#FF6800}{\dfrac {2x ^ {2} - 2x - 3} {x ^ {2} - 1}} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {2x ^ {2} - 2x - 3} {x ^ {2} - 1} \right)$
$ $ Factorize $ x ^ {2} $ in the formula $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {x ^ {2} \times (2 - \dfrac {2} {x} - \dfrac {3} {x ^ {2}})} {x ^ {2} - 1} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {x ^ {2} \times (2 - \dfrac {2} {x} - \dfrac {3} {x ^ {2}})} {x ^ {2} - 1} \right)$
$ $ Factorize $ x ^ {2} $ in the formula $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {x ^ {2} \times (2 - \dfrac {2} {x} - \dfrac {3} {x ^ {2}})} {x ^ {2} \times (1 - \dfrac {1} {x ^ {2}})} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {\color{#FF6800}{x ^ {2}} \times (2 - \dfrac {2} {x} - \dfrac {3} {x ^ {2}})} {\color{#FF6800}{x ^ {2}} \times (1 - \dfrac {1} {x ^ {2}})} \right)$
$ $ Reduce the fraction with $ x ^ {2} $ $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {2 - \dfrac {2} {x} - \dfrac {3} {x ^ {2}}} {1 - \dfrac {1} {x ^ {2}}} \right)$
$\color{#FF6800}{ \lim\limits_{x\to + ∞} } \left( \color{#FF6800}{\dfrac {2 - \dfrac {2} {x} - \dfrac {3} {x ^ {2}}} {1 - \dfrac {1} {x ^ {2}}}} \right)$
$ $ Calculate the limiting value $ $
$\color{#FF6800}{\dfrac {2 - 2 \times 0 - 3 \times 0} {1 - 0}}$
$\color{#FF6800}{\dfrac {2 - 2 \times 0 - 3 \times 0} {1 - 0}}$
$ $ Solve the formula $ $
$\color{#FF6800}{2}$
$2$
using L'Hospital's theorem
$\lim\limits_{x\to + ∞} \left( \dfrac {2x ^ {2} - 2x - 3} {x ^ {2} - 1} \right)$
$ $ Since the limit value of the numerator and denominator can be an indeterminate form, utilize L'Hospital's theorem $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {\dfrac{d}{dx}(2x ^ {2} - 2x - 3)} {\dfrac{d}{dx}(x ^ {2} - 1)} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {\dfrac{d}{dx}(2x ^ {2} - 2x - 3)} {\dfrac{d}{dx}(x ^ {2} - 1)} \right)$
$ $ Calculate the differentiation $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {4x - 2} {\dfrac{d}{dx}(x ^ {2} - 1)} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {4x - 2} {\dfrac{d}{dx}(x ^ {2} - 1)} \right)$
$ $ Calculate the differentiation $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {4x - 2} {2x} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {4x - 2} {2x} \right)$
$ $ Factorize $ 2 $ in the formula $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {2(2x - 1)} {2x} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {\color{#FF6800}{2}(2x - 1)} {\color{#FF6800}{2}x} \right)$
$ $ Reduce the fraction with $ 2 $ $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {2x - 1} {x} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {2x - 1} {x} \right)$
$ $ Since the limit value of the numerator and denominator can be an indeterminate form, utilize L'Hospital's theorem $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {\dfrac{d}{dx}(2x - 1)} {\dfrac{d}{dx}(x)} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {\dfrac{d}{dx}(2x - 1)} {\dfrac{d}{dx}(x)} \right)$
$ $ Calculate the differentiation $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {2} {\dfrac{d}{dx}(x)} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {2} {\dfrac{d}{dx}(x)} \right)$
$ $ Calculate the differentiation $ $
$\lim\limits_{x\to + ∞} \left( \dfrac {2} {1} \right)$
$\lim\limits_{x\to + ∞} \left( \dfrac {2} {1} \right)$
$ $ Dividing a particular formula by $ 1 $ , the formula remains unchanged $ $
$\lim\limits_{x\to + ∞} \left( 2 \right)$
$\color{#FF6800}{ \lim\limits_{x\to + ∞} } \left( \color{#FF6800}{2} \right)$
$ $ The value of limit is the same as the constant $ $
$\color{#FF6800}{2}$
Solution search results
search-thumbnail-$The\ reciprocal\\ 0+11) \left(\frac{2}
$8$ $\left(1$ Point) $1\right)$ The\ reciprocal\\ $0+11\right)$ \left(\frac{2} $c\left(2\right)$ {5}\right)^0\ $\right)$ \ $1111s\right)$ $S$ $S1S$ $s3S$ $S4S$ $s2S$
7th-9th grade
Other
Search count: 4,895
Check solutionright-arrow-icon
search-thumbnail-Using the \emph{removal of first derivative}
Using the \emph{removal of first derivative} method, the differential equation \( \frac{d^{2}y} $\left(d\times n$ $\left(2\right)\right)+P|ffac\left(dy\right)\left(dx\right)+Qy=F$ $dx\right)+Qy=RN\right)$ is transformed as \). For, the differential equation \frac{d^{2}y} $\left(d^{n}\left(2\right)y\right)$ $dx$ $\left(2\right)+2x$ $\left(0C\left(dy\right)\left(dx\right)+\left(x$ $2+1\right)y=\times n3+3x\right)$ the value of $\left(11\right)$
Calculus
Search count: 3,465
Check solutionright-arrow-icon
search-thumbnail-1+x x=\sqrt{(Nfrac(2ab)(a}+b) 1) then the value of )
Question $14$ Not yet answered Marked out of $1.00$ $1+x$ $x=\sqrt{\left(Nfrac\left(2ab\right)\left(a} +b\right)$ $1\right)$ then the value of $\right)$ $\left(fr0c\left(x+a\right)\left(x-a\right)+lfrac\left(x+b\right)\left(x-b\right)$ $1\right)$ $1s$ Select $One$ a. $a/b$ b. $1$ $○$ $c2$ $2$ $○$ d. $a^{2}+b^{2}$ $○$ CLEAR MY CHOICE
Other
Check solutionright-arrow-icon
search-thumbnail-Theis the statement of the Mean Value Theorem from your te\timestb00k
The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$
Calculus
Search count: 3,278
Check solutionright-arrow-icon
search-thumbnail-s|ef(-1n }\right)^{50}\ ) \ is\ equal\ to\ S
$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ $s1S$ $S-1S$ $s2S$ $s50s$
7th-9th grade
Other
Search count: 625
Check solutionright-arrow-icon
search-thumbnail-If the sum of two consecutive
If the sum of two consecutive numbers is $45$ and one number is $X$ .This statement in the form of equation $1s:$ $\left(1$ Point) $\right)$ $○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ $○sx+1ef\left(x+2$ $r1gnt\right)=145s$ $sx+1x=45s$
7th-9th grade
Algebra
Check solutionright-arrow-icon
Have you found the solution you wanted?
Try again
Try more features at Qanda!
check-iconSearch by problem image
check-iconAsk 1:1 question to TOP class teachers
check-iconAI recommend problems and video lecture
apple-logoApp Storegoogle-play-logoGoogle Play
© 2021 Mathpresso Inc.
|
CEO Jongheun Lee, Yongjae Lee
|
17th Floor, WeWork Seolleung Station III, 428, Seolleung-ro, Gangnam-gu, Seoul
|
EMAIL support.en@mathpresso.com