In the neighbourhood of `x=0` it is known that
`1+|x|lt(e^(x)-1)/(x)lt1-|x|"then find"lim_(xto0)(e^(x)-1)/(x).`
1 Answers
Talk Doctor Online in Bissoy App
Let `f(x)={{:(cos[x]", "xle0),(|x|+a", "xlt0):}.` Then find the value of a, so that `lim_(xto0) f(x)` exists, where [x] denotes the greatest integer f
Since `underset(xto0)limf(x)` exists, we have `underset(xto0-)limf(x)=underset(xto0+)limf(x)` or`""underset(hto0)limf(0-h)=underset(hto0)limf(0+h)` or`" "underset(hto0)lim||0-h||+a=underset(hto0)cos[0+h]` or`" "a=costheta=1` `:." "a=1`
2 Answers 1 views
Find the following limits. `(i)" "lim_(xto2) (4x)/(x^(3)-3)" "(ii)" "lim_(xto1)(log_(10)x-3)/(3x-2)" "(iii)" "lim_(xto pi) (3+cosx)/(2-sinx)`
In each of the given limits function is defined and contin-uously exists in the NBD of x where we have to find the limit. So, we will find the limits...
2 Answers 1 views
Find the following limits: `(i) lim_(xto0) (1-x)^((1)/(x))" "(ii) lim_(xto1) (1+log_(e)x)^((1)/(log_(e)x))` `(iii)lim_(xto0) (1+sinx)^((1)/(x))`
`(i) underset(xto)lim(1-x)^((1)/(x))=(underset(xto0)lim(1-x)^((1)/(-x)))^(-1)=e^(-1)` (ii) Since, `underset(xto1)log_(e)x=0, underset(xto1)lim(1+log_(e)x)^((1)/(log_(e)x))=1` `(iii)underset(xto0)lim (1+sinx)^((1)/(x))=underset(xto0)lim((1+sinx)^((1)/(sinx)))^((sinx)/(x))` `=(underset(xto0)lim(1+sinx)^((1)/(sinx)))^(underset(xto0)lim^((sinx)/(x)))` `=e^(1)=e`
2 Answers 1 views
If `f(x)=(tanx)/(x)`, then find `lim_(xto0)([f(x)]+x^(2))^((1)/([f(x)]))`, where `[.]` and `{.}` denotes greatest integer and fractional part function
We know that `underset(xto0)lim(tanx)/(x)=1^(+)` `:." "underset(xto0)lim[(tanx)/(x)]=1` and `underset(xto0)lim{(tanx)/(x)}=underset(xto0)lim((tanx)/(x)-[(tanx)/(x)])` `=underset(xto0)lim((tanx)/(x)-1)` `:." "underset(xto0)lim([f(x)]+x^(2))^((1)/({f(s)}))=underset(xto0)lim(1+x^(2))^(underset(xto0)lim((1)/(tanx))/(x)-1)` `=e^(underset(xto0)lim((x^(2))/(tanx))/(x)-1)` `=e^(underset(xto0)lim(x^(3)cosx)/(sinx-xcosx))` `=e^(underset(xto0)lim(x^(3))/((x+(x^(3))/(3!)+...)-x(1-(x^(2))/(2!)+...)))` `=e^(underset(xto0)lim(x^(3))/(-(x^(3))/(3!)+(x^(3))/(2!)))=e^(3)`
2 Answers 1 views
If `f(x)={{:(sinx","" "xnenpi", "ninI),(2","" ""otherwise"):}` and `g(x)={{:(x^(2)+1","" "xne0", "2),(4","" "x=0),(5","" "x=2):}` then find `lim_(xto0
Correct Answer - 1 `underset(xto0^(+))limg{f(x)}=g(f(0^(+)))=g((sin0^(+)))=g(0^(+))=(0)^(2)+1=1` `underset(xto0^(-))limg{f(x)}=g(f(0^(-)))=g((sin0^(-)))=g(0^(-))=(0)^(2)+1=1`
2 Answers 1 views
Find the values of a and b in order that `lim_(xto0) (x(1+acosx)-bsinx)/(x^(3))=1.`
Correct Answer - `a=(-5)/(2),b=(-3)/(2)` `underset(xto0)lim(x(1+acosx)-bsinx)/(x^(3))=1" "`(0/0 form) `impliesunderset(xto0)lim(x+axcosx-bsinx)/(x^(3))=1` `impliesunderset(xto0)lim(x+ax(1-(x^(2))/(2!)+(x^(4))/(4!)+...)-b(x-(x^(3))/(3!)+(x^(5))/(5!)+...))/(x^(3))=1` `impliesunderset(xto0)lim((1+a-b)x+(-(a)/(2)+(b)/(6))x^(3))/(x^(3))=1` `implies1+a-b=0" "(1)` and `-a/2+b/6=1" "(2)` Solving equations (1) and (2), we get `a=-(5)/(2),b=-(3)/(2).`
2 Answers 1 views
Evaluate`lim_(xto0^(+)) x^(x)"and "lim_(xto0^(+))x^(x^(x))`
Correct Answer - `1" and "0` Let`y=x^(x)` `:.logy=xlogx` `:.underset(yto0)limlogy=underset(xto0)limxlogx` `=underset(xto0)lim(logx)/(1//x)" "((oo)/(oo)" form")` `=underset(xto0)lim(1//x)/(-1//x^(2))` `=underset(xto0)lim(logx)/(1//x)" "((oo)/(oo)" form")` `=underset(xto0)lim(1//x)/(-1//x^(2))` `:." "y=1` `=underset(xto0^(+))limx^(x^(x))=(0^(+))^(1)=0`
2 Answers 1 views
If `lim_(xto0) (f(x))/(sin^(2)x)=8,lim_(xto0) (g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda" and "` `lim_(xto0) (1+2f(x))^((1)/(g(x)))=(1)/(e)," then"` The va
Correct Answer - C `underset(xto0)lim(f(x))/(sin^(2)x)=8` `implies" "underset(xto0)lim((f(x))/(x^(2)))/((sin^(2)x)/(x^(2)))=8` `implies" "underset(xto0)lim(f(x))/(x^(2))=8" "...(1)` Also, `underset(xto0)lim(g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda` `implies" "underset(xto0)lim((g(x))/(x^(2)))/((-2(1-cosx))/(x^(2))-(e^(x)-1)/(x)+x)=lamda` `implies" "underset(xto0)lim((g(x))/(x^(2)))/((-4"sin"^(2)(x)/(2))/(x^(2))=-1+0)=lamda` `implies" "underset(xto0)lim(g(x))/(x^(2))=-2lamda" "...(2)` Now, `underset(xto0)lim(1+2f(x))^((1)/(g(x)))`(one power infinity form) `=e^(underset(xto0)lim(2f(x))/(g(x)))=e^(underset(xto0)lim(2(f(x)//x^(2)))/((g(x)//x^(2))))` `=e^(underset(xto0)lim(16)/(-2lamda))` `=e^(-(8)/(lamda))` `e^(-1)`(given) `:." "lamda=8` `underset(xto0)lim(1+f(x))^((1)/(2g(x)))=e^(underset(xto0)lim(f(x))/(2g(x)))`...
2 Answers 1 views
If `lim_(xto0)(f(x))/(sin^(2)x)=8,lim_(xto0)(g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda" and "` `lim_(x to 0)(1+2f(x))^((1)/(g(x)))=(1)/(e)," then"` `lim_(x
Correct Answer - A `underset(xto0)lim(f(x))/(sin^(2)x)=8` `implies" "underset(xto0)lim((f(x))/(x^(2)))/((sin^(2)x)/(x^(2)))=8` `implies" "underset(xto0)lim(f(x))/(x^(2))=8" "...(1)` Also, `underset(xto0)lim(g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda` `implies" "underset(xto0)lim((g(x))/(x^(2)))/((-2(1-cosx))/(x^(2))-(e^(x)-1)/(x)+x)=lamda` `implies" "underset(xto0)lim((g(x))/(x^(2)))/((-4"sin"^(2)(x)/(2))/(x^(2))=-1+0)=lamda` `implies" "underset(xto0)lim(g(x))/(x^(2))=-2lamda" "...(2)` Now, `underset(xto0)lim(1+2f(x))^((1)/(g(x)))`(one power infinity form) `=e^(underset(xto0)lim(2f(x))/(g(x)))=e^(underset(xto0)lim(2(f(x)//x^(2)))/((g(x)//x^(2))))` `=e^(underset(xto0)lim(16)/(-2lamda))` `=e^(-(8)/(lamda))` `e^(-1)`(given) `:." "lamda=8` `underset(xto0)lim(1+f(x))^((1)/(2g(x)))=e^(underset(xto0)lim(f(x))/(2g(x)))`...
2 Answers 1 views
If `lim_(xtooo) f(x)` exists and is finite and nonzero and if `lim_(xtooo) {f(x)+(3f(x)-1)/(f^(2)(x))}=3`, then the value of `lim_(xtooo) f(x)" is "`_
Correct Answer - `(1)` `underset(xtooo)lim(f(x)+(3f(x)-1)/(f^(2)(x)))=3` or `(underset(xtooo)limf(x)+(3underset(xtooo)limf(x)-1)/((underset(xtooo)limf(x))^(2)))=3` or `(y+(3y-1)/(y^(2)))=3` or `y^(3)-3y^(2)+3y-1=0` or `(y-1)^(3)=0` or `y=1`
2 Answers 1 views