# Homework Help

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• Math
Let's transform the expression to get a form valid for use of l’Hospital’s Rule. `x^(sqrt(x)) = (e^(lnx))^(sqrt(x)) = e^(sqrt(x)*lnx) = e^((lnx/(x^(-1/2))))` Consider the power,...

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• Math
When `x->0^+,` 2x also `->0^+.` And tan(y) is continuous at 0 as an elementary function inside its domain. Therefore `lim_(x->0^+)(tan(2x)) = tan(0)` = 0. l’Hospital’s Rule isn't...

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• Math
You need to evaluate the given limit, hence, you need to replace 0 for x, such that: `lim_(x->0)(1 - 2x)^(1/x) = (1 - 0)^(1/0) = 1^oo` You may use special limit `lim_(x->0) (1 + x)^(1/x) = e`...

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• Math
We will use the fact that `lim_(x->oo)(1+1/x)^x=e.` `lim_(x->oo)(1+a/x)^(bx)=lim_(x->oo)(1+a/x)^(bx/a cdot a)` Now we use substitution `y=x/a.` `=lim_(y->oo)(1+1/y)^(aby)`...

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• Math
You need to evaluate the limit, using logarithm special technique, such that: `f(x) = x^(1/(1-x))ln f(x) = ln (x^(1/(1-x)))` `ln f(x) = 1/(1-x)*ln x` `lim_(x->1^+) 1/(1-x)*ln x = 0/0` Since the...

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• Math
Given the limit function `lim_{x->infty}x^((ln(2)/(1+ln(x))` . We have to find the limit value. This is of the form `infty^0` and can be written as:...

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• Math
Given the limit function `lim_{x->0}(sinh(x)-x)/x^3` `` Applying the limits we have, `lim_{x->0}(sinh(x)-x)/(x^3)=0/0` Using L'Hospital's rule we get, `lim_{x->0}(cosh(x)-1)/(3x^2)=0/0`...

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• Math
First transform the expression: `tanh(x)/tan(x) = (sinh(x)/cosh(x)) / (sin(x)/cos(x)) = (sinh(x)/sin(x))*(cos(x)/cosh(x)).` The second fraction obviously has limit 1 when `x->0`, and we can omit...

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• Math
`lim_(x->0)(x-sin x)/(x-tan x)` Apply L'Hospital's rule. `=lim_(x->0)(1-cos x)/(1-1/cos^2x)` Apply L'Hospital's rule again. `=lim_(x->0)sin x/(-(2sin x)/cos^3x)` Cancel `sin x.`...

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• Math
Given the limit `lim_{x->0}sin^(-1)(x/x)` . We have to find the limit. Applying the limit we get, `lim_{x->0}sin^(-1)(x/x)=lim_{x->0}sin^(-1)(1)=pi/2` hence the limit is `pi/2`

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• Math
You need to evaluate the limit, such that: `lim_(x->oo) (ln^2 x)/x= oo/oo` Since the limit is indeterminate `oo/oo` , you may use l'Hospital's rule: `lim_(x->oo) ((ln^2 x)')/(x') =...

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• Math
`lim_(x->0)(x3^x)/(3^x-1)` Apply L'Hospital rule , Test condition:0/0 `=lim_(x->0)((x3^x)')/((3^x-1)')` `=lim_(x->0)(x3^xln(3)+3^x)/(3^xln(3))` `=lim_(x->0)(3^x(xln(3)+1))/(3^xln(3))`...

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• Math
Given the limit `lim_{x->0}(cos(mx)-cos(nx))/x^2` . We have to find the limit value Applying the limits we get, `lim_{x->0}(cos(mx)-cos(nx))/x^2=0/0` Using L'Hospital's rule and then applying...

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• Math
`lim_(x->0)(x+sin(x))/(x+cos(x))` plug in the value, `=(0+sin(0))/(0+cos(0))` `=(0+0)/(0+1)` `=0/1` =0

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• Math
Given the limit `lim_{x->0}x/(tan^(-1)(4x))` . We have to find the limit value. Applying the limit we get, `lim_{x->0}x/(tan^(-1)(4x))=0/0` So we use L'Hospital's rule to obtain,...

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• Math
`lim_(x->1) (1-x+ln(x))/(1+cos(x))` To compute for its limit, plug-in x=1. `= (1-1+ln(1))/(1+cos(1))` `=0/(1+cos(1))` `=0` The result is finite value. This means that the function is defined at...

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• Math
The denominator has limit `-oo+0-1=-oo.` The numerator has limit 0 (I'll prove this below). Therefore there is no indeterminacy, `0/oo=0.` (zero is the answer). Now prove that `x^x->1` when...

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• Math
Given the limit `lim_{x->1}(x^a-ax+a-1)/(x-1)^2` . We have to find the limit value. Applying the limits we get, `lim_{x->1}=(x^a-ax+a-1)/(x-1)^2=(1^a-a+a-1)/(1-1)^2=0/0` Since `1^a=1` So...

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• Math
Given the limit function `lim_{x->0}(e^x-e^(-x)-2x)/(x-sin(x))` . We have to find the limits. Applying the limits we get, `lim_{x->0}(e^x-e^(-x)-2x)/(x-sin(x))=0/0` So we have to apply the...

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• Math
Given limit function is: `lim_{x->0}(cos(x)-1+(1/2)x^2)/x^4` We have to find the limits. Applying the limit we see that, `lim_{x->0}(cos(x)-1+(1/2)x^2)/x^4=0/0` So now applying the...

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• Math
You need to evaluate the limit, hence, you need to replace `a^+` for x in limit, such that: `lim_(x->a^+) (cos x*ln (x - a))/(x - sin x) = (cos a*ln (a - a))/(a- sin a)` Notice that `ln(a - a)...

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• Math
`lim_(x->oo)xsin(pi/x)` `=lim_(x->oo)sin(pi/x)/(1/x)` Apply L'Hospital rule , Test condition:0/0 `=lim_(x->oo)(sin(pi/x)')/((1/x)')` `=lim_(x->oo)(cos(pi/x)(-pix^-2))/(-1x^-2)`...

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• Math
As `x->+oo,` `sqrt(x) -> +oo` and `e^(-x/2) -> 0.` So the initial expression is an indeterminate form `oo*0.` Let's transform it to use l’Hospital’s Rule: `sqrt(x)*e^(-x/2) =...

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• Math
`lim_(x->0) cot(2x)sin(6x)` The function cot(2x)sin(6x) is undefined at x=0. So to take its limit, let's apply the L'Hospital's Rule. To do so, express it as a rational function. `=lim_(x->0)...

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• Math
You need to evaluate the limit such that: `lim_(x->0^+) sin x*ln x = 0*(-oo)` You need to use the special limit `lim_(x->0^+) (sin x)/x = 1` , such that: `lim_(x->0^+) ((sin x)/x)*x*ln x=...

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• Math
`lim_(x->1/2)(6x^2+5x-4)/(4x^2+16x-9)` `=lim_(x->1/2)((2x-1)(3x+4))/((2x-1)(2x+9))` `=lim_(x->1/2)(3x+4)/(2x+9)` plug in the value, `=(3*1/2+4)/(2*1/2+9)` `=(3+8)/(2+18)=11/20`

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• Math
You need to evaluate the limit, hence, you need to replace `pi/2 ` for x such that: `lim_(x->pi/2) (cos x)/(1 - sin x) = (cos (pi/2))/(1 - sin (pi/2)) = 0/(1-1) = 0/0` Since the limit is...

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• Math
Given the limit function `lim_{x->0}sin(4x)/tan(5x)` . We have to find the limit. Applying the limit we can see that, `lim_{x->0}sin(4x)/tan(5x)=0/0` which is of the form `0/0` . Hence we...

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• Math
You need to evaluate the limit, hence, you need to replace 0 for t in the limit, such that: `lim_(t->0) (e^(2t) - 1)/(sin t) = (e^0 - 1)/(sin 0) = (1-1)/0 = 0/0` Since the limit is indeterminate...

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• Math
`lim_(x->0) x^2/(1-cos(x))` The function `f(x) = x^2/(1-cosx)` is undefined at x=0. So to compute its limit as x approaches zero, apply the L'Hospital's Rule. Take the derivative of the...

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• Math
You need to evaluate the limit, hence, you need to replace pi/2 for theta in limit, such that: `lim_(theta->pi/2) (1 - sin theta)/(1 + cos 2theta) = (1 - sin(pi/2))/(1 + cos pi)`...

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• Math
`lim_(theta->pi/2) (1-sin theta)/csc theta` The function `(1-sin theta)/csc theta` is defined at `theta=pi/2` . So to take its limit, there is no need to apply the L'Hospital's Rule. Instead,...

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• Math
`lim_(x->oo)ln(x)/sqrt(x)` Apply L'Hospital rule , Test L'Hospital condition:`oo/oo` `=lim_(x->oo)((ln(x))')/((sqrt(x))')` `=lim_(x->oo)(1/x)/((1/2)(x^(-1/2)))` `=lim_(x->oo)2/sqrt(x)`...

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• Math
`lim_(x->oo)(x+x^2)/(1-2x^2)` `lim_(x->oo)(x^2(1/x+1))/(x^2(1/x^2-2))` `lim_(x->oo)(1/x+1)/(1/x^2-2)` apply infinity properties, `=(0+1)/(0-2)=-1/2`

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• Math
You need to evaluate the given limit, such that: `lim_(x->0^+) (ln x)/x = (ln 0^+)/(0^+)` `lim_(x->0^+) (ln x)/x = (-oo)*1/(0^+)` `lim_(x->0^+) (ln x)/x = -oo*(+oo) = -oo` Hence,...

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• Math
`lim_(x->oo)(lnsqrt(x))/x^2` Apply L'Hospital rule , Test L'Hospital condition:`oo/oo` `=lim_(x->oo)((lnsqrt(x))')/((x^2)')` `=lim_(x->oo)((1/sqrt(x))(1/2)(x^(-1/2)))/(2x)`...

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• Math
`lim_(t->1)(t^8-1)/(t^5-1)` `=lim_(t->1)((t-1)(t+1)(t^2+1)(t^4+1))/((t-1)(t^4+t^3+t^2+t+1))` `lim_(t->1) ((t+1)(t^2+1)(t^4+1))/((t^4+t^3+t^2+t+1))` plug in the value,...

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• Math
`lim_(t->oo)(8^t-5^t)/t` Apply L'Hospital rule, Test condition:0/0 `=lim_(t->oo)((8^t-5^t)')/(t')` `=lim_(t->oo)(8^tln(8)-5^tln(5))` plug in the value, `=8^0ln(8)-5^0ln(5)` `=ln(8)-ln(5)`...

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• Math
`lim_(x->0)(sqrt(1+2x)-sqrt(1-4x))/x` `=lim_(x->0)((sqrt(1+2x)-sqrt(1-4x))(sqrt(1+2x)+sqrt(1-4x)))/(x(sqrt(1+2x)+sqrt(1-4x)))` `=lim_(x->0)((1+2x)-(1-4x))/(x(sqrt(1+2x)+sqrt(1-4x)))`...

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• Math
You need to evaluate the limit, hence, you need to replace `oo` for u: `lim_(u->oo) (e^(u/10))/(u^3) = (e^oo)/(oo) = oo/oo` Since the limit is indeterminate `oo/oo` , you may apply l'Hospital's...

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• Math
Given the limit function `lim_{x->0}(e^x-1-x)/x^2` . We have to find the limit. We can see that the above limit is of the form `0/0` . So we have to apply L'Hospital's rule. i.e. differentiating...

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• Math
`lim_(x->1)(x^2-1)/(x^2-x)` `=lim_(x->1)((x+1)(x-1))/(x(x-1))` `=lim_(x->1)(x+1)/x` plug in the value, `=(1+1)/1` =2

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• Math
`lim_(x->2)(x^2+x-6)/(x-2)` `=lim_(x->2)((x-2)(x+3))/(x-2)` `=lim_(x->2)(x+3)` plug in the value, =(2+3) =5

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• Math
`lim_(x->1)(x^3-2x^2+1)/(x^3-1)` `=lim_(x->1)((x-1)(x^2-x-1))/((x-1)(x^2+x+1))` `=lim_(x->1)(x^2-x-1)/(x^2+x+1)` plug in the value, `=(1^2-1-1)/(1^2+1+1)` =-1/3

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• Science
The transition metals are the metals that make up groups 3-12 of the periodic table. This region is called the d-block because it corresponds to the d electron sublevels filling. Some of the bettor...

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• Law and Politics
One of the main reasons why the Southern states seceded was that they wanted states to have more power (so they could have slavery). To make sure that the states had more power, they made a new...

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• History
The period of "New Immigrants" between 1885-1920 marked a shift in the type of people coming to the United States. Immigrants from this era were primarily from south and east Europe. Immigration...

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• The Great Gatsby
Nick's comment about Gatsby's worth is an accurate one in that he actually earned all of his money, albeit through illegal means. "The bunch" Nick refers to here—Daisy and Tom—inherited their...

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• A Christmas Carol
Jacob Marley had the greatest effect on Scrooge, because he created the opportunity for his transformation. There was not just one character who had an influence on Scrooge. However, since Jacob...

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