1. Find ltx → ∞\((1+\frac{1}{x^2+2x+1})^{x^2+3x+1}\)
a) e
b) 1
c) e2
d) 1⁄e
Explanation: Use the form
lt x→ ∞(1 + f(x))g(x) = elt x→ ∞ f(x) * g(x)
Provided as x → ∞ we must have
f(x) → 0
g(x) → ∞
These conditions are met in our question
L = ltx → ∞\((1+\frac{1}{x^2+2x+1})^{x^2+3x+1} = e^{\frac{x^2+3x+1}{x^2+2x+1}}\)
ltx → 0\(\frac{x^2+3x+1}{x^2+2x+1}=1\)
L = e1 = e.
2. Find lt x → ∞\((\frac{ln(1+\frac{(x+3)^3(2x+9)}{(4x^3+3)})}{x^3+3x^2+9x+27})\)
a) 0
b) 1
c) Undefined
d) – 1⁄35
Explanation: The form here is of 0⁄0
Applying L hospitals rule would be really tough to differentiate. Hence we use the concept of Taylor Series
We know that ln(1+x)=\(x-\frac{x^2}{2}+\frac{x^3}{3}-…\infty\)
Thus, we have
\(=lt_{x\rightarrow -3}(\frac{1}{x^3+3x^2+9x+27}\times (\frac{(x+3)^3(2x+9)}{4x^3+3}-\frac{(x+3)^6(2x+9)^2}{2(4x^3+3)^2}+..\infty))\)
\(=lt_{x\rightarrow -3}(\frac{1}{x+3}^3\times(\frac{(x+3)^3(2x+9)}{4x^3+3}-\frac{(x+3)^6(2x+9)^2}{2(4x^3+3)^2}+..\infty))\)
\(=lt_{x\rightarrow -3}(\frac{(2x+9)}{4x^3+3}-\frac{(x+3)^3(2x+9)^2}{2(4x^3+3)^2}+…\infty)\)
All the terms except the first one go to zero, we now have
\(=lt_{x\rightarrow -3}\frac{(2x+9)}{(4x^3+3)}=\frac{2(-3)+9}{4(-3)^3+3}=\frac{3}{-105}\)
\(=-\frac{1}{35}\)
3. Find ltn → ∞\(\sum_{a=0}^{n-1}\frac{sin(\frac{a}{n})}{n}\)
a) 1⁄a
b) 1
c) 1 – cos(1)
d) 0
Explanation: We use the concept of limit of a sum which is
\(\int_a^b f(x)dx=lt_{n\rightarrow \infty}(\frac{b-a}{n})\times(f(a)+f(a+\frac{b-a}{n})….+f(a+\frac{(n-1)(b-a)}{n}))\)
Thus we have
\(\int_0^1 sin(x)dx=lt_{n\rightarrow \infty}(\frac{1}{n})\times(f(0)+f(\frac{1}{n})+….+f(\frac{n-1}{n}))\)
\(=lt_{n\rightarrow\infty}\frac{1}{n} \times (sin(\frac{1}{n})+sin(\frac{2}{n})+…+sin(\frac{n-1}{n}))\)
\(\int_0^1 sin(x)dx=lt_{n\rightarrow \infty}\sum_{a=0}^{n-1}\frac{sin(\frac{a}{n})}{n}\)
It is enough to evaluate the integral
\(\int_{0}^{1} sin(x)dx=[-cos(x)]_0^1\)=(cos(0)-cos(1))
=(1-cos(1))
4. Find ltx → 0\((\frac{ln(1+x^4)}{x})\)
a) 1
b) -1
c) 0
d) Undefined
Explanation:
ltx → 0\((\frac{ln(1+x^4)}{x}) = lt_{x\rightarrow 0} (\frac{1}{x}) \times (\frac{x^4}{1}-\frac{x^8}{2}+\frac{x^{12}}{3}-…\infty)\)
ltx → 0\((\frac{x^3}{1}-\frac{x^7}{2}+\frac{x^{11}}{3}-…\infty)\)
= 0.
5. Find \(lt_{x\rightarrow 0}(\frac{1}{sin^2(x)})\)
a) 2
b) 1
c) 0
d) Undefined
Explanation:
=\(lt_{x\rightarrow 0}(\frac{1}{(\frac{1-cos(2x)}{2})})=lt_{x\rightarrow 0}(\frac{2}{1-cos(2x)})\)
=\((\frac{2}{1-cos(0)})=\frac{2}{0} \rightarrow \infty\)
6. Find \(lt_{x\rightarrow \infty}((\frac{x^3+x^2+x}{x^3+x+1})^{x+3})\)
a) e
b) e-1
c) 0
d) 1
Explanation:
=\(lt_{x\rightarrow\infty}(1+\frac{x^2-1}{x^3+x+1^{x+3}})^{x+3}=e^{lt_{x\rightarrow\infty}(\frac{(x^2-1)(x+3)}{x^3+x+1})}\)
\(lt_{x\rightarrow\infty}(\frac{(x^2-1)(x+3)}{x^3+x+1})=1\)
= e1 = e.
7. Find \(lt_{x\rightarrow -2}(\frac{sin((x-2)^2)}{(x+2)^2})\)
a) 1
b) 0
c) ∞
d) 0⁄0
Explanation: We have 0⁄0 form
Now we have the form
\(lt_{x\rightarrow a}\frac{sin(f(x))}{g(x)}\)=1
where f(x) → 0:g(x) → 0 as x → a
∴\(lt_{x\rightarrow -2}(\frac{sin((x-2)^2)}{(x-2)^2})=1\)
8. Find \(lt_{n\rightarrow\infty}\sum_{a=1}^{n-1}(\frac{ln(1+\frac{a}{n})}{n})\)
a) ln(2)
b) ln(4)
c) 3ln(2)
d) 1⁄a
Explanation: Using limit of sum we have
=\(lt_{n\rightarrow\infty}(\frac{1}{n})\times(ln(1+0)+ln(1+\frac{1}{n})+ln(1+\frac{2}{n})+…+ln(1+\frac{(n-1)}{n}))\)
=\(\int_0^1 ln(1+x)dr=[(x+1)(ln(x+1)-1)]_0^1\)
=(2ln(2)-1)-(ln(1)-1)=2ln(2)
=ln(4)
9. Find \(lt_{n\rightarrow\infty}\frac{(1^a+2^a+…+(n-1)^a)}{n^{a+1}}\)
a) 1
b) 1⁄a + 1
c) 0
d) Undefined
Explanation: \(\int_0^1x^a dx=lt_{n\rightarrow\infty}\frac{1}{n}\times((\frac{1}{n})^a+(\frac{2}{n})^a+…+(\frac{n-1}{n})^a)\)
=\(lt_{n\rightarrow\infty}\frac{(1^a+2^a+…+(n-1)^a)}{n^{a+1}}\)
It is enough to evaluatethis integral
\(\int_0^1 x^adx=[\frac{x^{a+1}}{a+1}]_0^1\)
=\(\frac{1}{a+1}\)
10. Find \(lt_{x\rightarrow -101}(\frac{ln(x^2+20x+(x+101)^2(x^2+3))-ln(x^2+20x)}{x^2+202x+10201})\).
a) 0
b) 1
c) ∞
d) \(=\frac{10204}{12221}\)
Explanation: \(=lt_{x\rightarrow -101}(\frac{ln(1+\frac{(x+101)^2(x^2+3)}{(x^2+20x)}}{(x+101)^2})\)
Now expanding into Taylor series we have
\(=lt_{x\rightarrow -101}(\frac{1}{(x+101)^2})\times(\frac{(x+101)^2(x^2+3)}{(x^2+20x)}-\frac{(x+101)^4(x^2+3)^2}{2(x^2+20x)^2}+…\infty)\)
\(=lt_{x\rightarrow -101}(\frac{(x^2+3)}{(x^2+20x)}-\frac{(x+101)^2(x^2+3)^2}{2(x^2+20x)^2}+….\infty)\)
All others exceptthe first term tend to zero. Thus, we have
\(=lt_{x\rightarrow -101}(\frac{(-101)^2+3}{(101)^2+20(101)})\)
\(=\frac{10204}{12221}\)