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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
`lim _(xto pi//2)(-2x+pi)/(cos x)` का मान होगा-A. 1B. `-2`C. 2D. 0 |
| Answer» Correct Answer - C | |
| 2. |
मान लीजिये `f,RtoR` एक धन वर्धमान फलन है जिसमे `lim_(xtooo)(f(3x))/(f(x))=1` है, तो `lim_(xtooo)(f(2x))/(f(x))` बराबर है-A. 1B. `2//3`C. `3//2`D. 3 |
| Answer» Correct Answer - A | |
| 3. |
परिगणना कीजिए `lim_(xto0)((3^(x)-2^(x))/(x))` |
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Answer» प्रश्नानुसार `lim_(xto0)((3^(x)-2^(x))/(x))=lim_(xto0){((3^(x)-1)-(2^(x)-1))/(x)}` `=lim_(xto0)((3x^(2)-1)/(x))-lim_(xto0)((2^(x)-1)/(x))` `=log3 -log2" "[lim_(xtoo)(a^(x)-1)/(x)=log_(e)a`का प्रयोग करने पर] |
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| 4. |
परिगणना कीजिए `:lim_(xto2)((e^(x)-e^(2))/(x-2))` |
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Answer» प्रश्नानुसार `lim_(xto2)((e^(x)-e^(2))/(x-2))=lim_(y to0)((e^(y+2)-e^(2))/(y))" "(x-2=yimpliesxto2impliesyto0)` `=lim_(yto0)[e^(2)*((e^(y-1))/(y))]` `=e^(2)lim_(yto0)((e^(y)-1)/(y))` `=e^(2)xx1=e^(2)" "[becauselim_(xto0)((e^(y)-1)/(y))=1]` |
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| 5. |
परिगणना कीजिए `"lim_(xto0)((sin 2 x+sin 6x)/(sin 5x-sin 3x))` |
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Answer» प्रश्नानुसार `lim_(xto0)((sin 2x+sin 6x)/(sin 5x-sin 3x))=lim_(xto0)((2sin4xcos 2x)/(2cos4x sinx))` `=lim_(xto0)((sin 4 x cos 2x)/(cos 4x sinx))` `=lim_(xto0)((sin4x)/(4x)xx(x)/(sinx)xx cos2x xx(1)/(cos 4xxx4)` `=4 lim_(xto0)((sin4x)/(4x))xxlim_(xto0)((x)/(sin x))xxlim_(xto0)cos 2 x xxlim_(xto0)((1)/(cos 4x))` `=4xx1xx1xx1xx1/1=4` |
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| 6. |
परिगणना कीजिए `lim_(xto0)((1-cos x)/(x^(2)))` |
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Answer» प्रश्नानुसार `lim_(xto0)((1-cosx)/(x^(2)))=lim_(xto0)(2sin^(2)x//2)/(((x)/(2))^(2)*4)" "(because cos x=1-2sin ^(2)""(x)/(2))` `=1/2lim_(xto0)[(sin ((x)/(2)))/((x)/(2))]^(2)=1/2xx1^(2)=1/2` |
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| 7. |
`lim_(x to 0) ((tan 2x-x)/(3x -sin x))` का मान ज्ञात कीजिए। |
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Answer» `underset(x to 0)lim ((tan 2x-x)/(3x -sin x))=underset(xto0)lim(((tan 2x)/(2x))2x-x)/(3x-((sin x)/(x))*x)` `=underset(x to 0)lim(((tan 2x)/(2x))2-1)/(3-(sin x)/(x))=(2*1-1)/(3-1)=1/2` |
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| 8. |
परिगणना कीजिए `:lim_(xto0)(x tan4x)/(1-cos 4x)` |
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Answer» प्रश्नानुसार, `lim_(xto0)(x tan 4x)/(1-cos 4x)lim_(xto0)(x sin4x)/(cos 4x(1-cos 4x))` ` =lim_(xto0)(2x sin2x cos2x)/(cos 4x(2sin^(2)2x))` `=lim_(xto0)((cos2x)/(cos 4x)*(2x)/(sin 2x))*1/2` `=1/2*(2xto0)/(lim_(4xto0)cos 4x)*lim_(2x to 0)((2x)/(sin 2x))` `=1/2 xx1xx1=1/2` |
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| 9. |
परिगणना कीजिए `:lim_(xto0)((1-cos 4x)/(1-cos5x))` |
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Answer» प्रश्नानुसार,`lim_(xto0)((1-cos4x)/(1-cos 5x))=lim_(xto0)(2sin^(2)2x)/(2sin^(2)((5x)/(2)))=lim_(xto0)(sin^(2)2x)/(sin^(2)((5x)/(2)))` `=lim_(xto0){(((sin2x)/(2x))^(2)*4x^(2))/({(sin((5x)/(2)))/((5x)/(2))}^(2)*(25x^(2))/(4))}=16/25{(lim_(2xto0)((sin2x)/(2x))^(2))/(lim_(5xto0)[(sin((5x)/(2)))/((5x)/(2))]^(2))}` `16/25xx(1^(2))/(1^(2))=16/25` |
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| 10. |
`lim_(x to 0) ((1-cos x cos2x cos3x)/(sin ^(2)2x))` का मान ज्ञात कीजिए। |
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Answer» हम जानते है कि `cos x cos2x cos 3x=1/2(2 cos x cos3x cos2x)` `=1/2[(cos 2x+cos 4x)cos2x]` `=1/4[(2 cos ^(2)2x+2 cos 4x cos 2x)]` `=1/4[1+cos 4x+cos 2x+cos 6x]` इसलिए `underset(x to0)lim[(1-cos x cos2x cos 3x)/(sin^(2)2x)]=underset(xto0)lim[(1-(1)/(4)(1+cos 4x+cos 2x+cos 6x))/(sin^(2)2x)]` `=underset(xto0)lim[(1-cos 2x+1-cos4x+1-cos 6x)/(4 sin^(2)2x)]` `=underset(x to 0)lim(2 sin ^(2)x+2sin ^(2)2x+2sin^(2)3x)/(4sin^(2)2x)` `=underset(x to 0)lim{(2((sin x)/(x))^(2)*x^(2)+2((sin 2x)/(2x))^(2)*4x^(2)+2((sin 3x)/(3x))^(2)*9x^(2))/(4((sin 2x)/(2x))*4x^(2))}` `=(28)/(16)=7/4` |
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| 11. |
`lim_(x to y) (tan x-tany)/(x-y)` का मान ज्ञात कीजिए। |
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Answer» `underset(x to y)lim(tan x-tan y)/(x-y)=underset(hto0)lim (tan (y+h)-tan y)/(y+h-y)` `underset(hto0)lim 1/h[sin (y+h)/(cos (y+h))-(sin y)/(cos y)]=underset(hto0)lim(sin (y+h)cos y-cos (y+h)sin y)/(hcos (y+h)cos y)` `=underset(hto0)lim(sin (y+h-y))/(h cos(y+h-y))` `=underset(hto0)lim(sin h)/(h)*(1)/(cos (y+h)cos y)=1*(1)/(cos ^(2)y)=sec^(2)y` |
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| 12. |
मान ज्ञात कीजिये- `lim_(xto0)(a^(x)-1)/(x),a gt0` |
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Answer» Correct Answer - `log_(e)a` `a^(x)=1+x log _(e)a+(x^(2))/(2!)(log_(e)a)^(2)+(x^(3))/(3!)(log_(e)a)^(3)+...` का प्रयोग करते हुए। |
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| 13. |
मान ज्ञात कीजिये- `lim_(xto0)(sin x)/(x+5)` |
| Answer» माना `x=1/y,` तब यदि `xtooo` तब `y to 0` | |
| 14. |
मान ज्ञात कीजिये- `lim_(xto0) (e^(x)-1)/(e^(x))` |
| Answer» `lim_(xto0)(e^(x)-1)/(e^(x))lim_(xto0)((1-(1)/(e^(x)))e^(x))/(e^(x))=lim_(xto0)(1-e^(-x))/(1)=1-e^(-0)=1-1=0` | |
| 15. |
`a in R` के लिए `a ne=1,` यदि `lim_(ntooo)(1^(a)+2^(a)+।।।+n^(a))/((n+1)^(a-1)[(na+1)+(na+2)+।।।+(na+n)])=1/60` तक a का मान ज्ञात कीजिए। |
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Answer» माना, `lim_(ntooo)(1^(a)+2^(a)+...+n^(a))/((n+1)^(a-1)[(na+1)+(na+2)+...+(na+n)])` `=lim_(xtooo)(n^(a)[((1)/(n))^(a)+((2)/(n))^(a)+...((n)/(n))^(a)])/((n+1)^(a-1)[n^(2)a+(n(n+1))/(2)])` `=lim_(xtooo)(n^(a)*n[1/n underset(r=1)overset(n)sum((r)/(a))^(a)])/((n+1)^(a-1). n(na+((n+1))/(2)))` `=lim_(xtooo)(n^(a))/((n+a)^(a-1))*((1)/(a+1))/(na+(n+1)/(2))` `=lim_(ntooo)(1)/((1+(1)/(n))^(a-1))*((1)/(a+1))/(a+(1)/(2)+(1)/(2n))` `=(2)/((a+1)(2a+1))=1/60` (दिया है) `implies(a+1)(2a+1)=120` ` impliesa=7, -17/2` |
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| 16. |
यदि फलन `f(x), lim_(xto1)(f(x)-2)/(x^(2)-1)=pi` को संतुष्ट करता है तो `lim_(xto1)f(x)` का मान ज्ञात कीजिए। |
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Answer» प्रश्नानुसार `lim_(xto1)(f(x)-2)/(x^(2)-1)=pi" "...(i)` स्पष्ट है कि `lim_(xto1)(x^(2)-1)=0` इसलिए यदि `lim_(xto1)f(x)-2ne0` तब `lim_(xto1)(f(x)2)/(x^(2)-1)=oo` या `-oo` जोकि सम्बन्ध (i ) से संभव नहीं है। इसलिए `lim_(xto1)[f(x)-2]=0` `implieslim_(xto1)f(x)=2` |
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| 17. |
यदि `f(x)={{:(a+bx,xlt1),(4:x=1),(b-ax,x gt 1):}` व `lim_(xto1)f(x)=f(1)` तब a व b के मान ज्ञात कीजिए। |
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Answer» प्रश्नानुसार `f(x)={{:(a+bx,xlt1),(4:x=1),(b-ax,x gt 1):}` `impliesf(1)=4" "...(i)` `lim_(xto1^(-))f(x)=lim_(xto1^(-))(a+bx)=a+b" "...(ii)` व `lim_(xto1^(+))f(x)=lim_(xto1^(+))(b-ax)=b-a" "(iii)` दिया है कि `lim_(xto1)f(x)` का अस्तित्व है इसलिए `lim_(xto1^(-))f(x)=lim_(xto1^(+))f(x)` `impliesa+b=b-a` `implies2a=0` `impliesa=0` इसी प्रकार `lim_(xto1)f(x)=f(1)` `a+b=4` `impliesb=4 " "(becausea=0)` अतः `a =0 ` तथा `b =4 ` |
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| 18. |
यदि `f(x)={{:(3-x^(2),x le-2),(ax+b,-2ltx lt2),((x^(2))/(2),xge2):},` तब `lim_(xto2)f(x)` तथा `lim_(xto-2)f(x)` के अस्तित्व के होने के लिए सिद्ध कीजिए कि `a=3/4, b=1/2` |
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Answer» `(lim_(xto2-0)f(x)=lim_(hto0)3-(-2-h)^(2)=-1)=(lim_(xto2+0)f(x)=lim_(hto0)a(-2+h)+b=-2a+b)` `implies" "-2a+b=-1" "…(i)` इसी प्रकार `x =2 ` पर, `" "2a+b=2" "…(ii)` अब समीकरण (i ) व (ii ) को हल करने पर, |
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| 19. |
`lim_(x tooo)(sqrt(x^(2)+x+1)-x)` का मान ज्ञात कीजिए। |
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Answer» `lim_(xtooo)(sqrt(x^(2)+x+1)-x)=lim_(xtooo)((sqrt(x^(2)+x+1)-x)(sqrt(x^(2)+1+x)+x))/((sqrt(x^(2)+x+1)+x))` `=lim_(xtooo)(x^(2)+x+1-x^(2))/(sqrt(x^(2)+x+1)+x)=lim_(xtooo)(x+1)/(sqrt(x^(2)+x+1)x+)` `=lim_(xtooo)(x+1)/(sqrt(1+(1)/(x)+(1)/(x^(2))))=lim_(x to oo)(x(1+(1)/(x)))/(x(sqrt(1+(1)/(x)+(1)/(x^(2))+1)))` `=lim_(xtooo)(1+(1)/(x))/(sqrt(1+(1)/(x)+(1)/(x^(2))+1))=1/2` |
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| 20. |
मान ज्ञात कीजिए- `lim_(xto-1)(1+x+x^(2)+...+x^(10))` |
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Answer» Correct Answer - 1 `lim_(xto-1)(1+x+x^(2)+...+x^(10))=1+(-1)+(-1)^(9)+(-1)^(10)=1-1-1+1-1-...(-1)+1=1` |
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| 21. |
यदि `alpha ` व `beta ` समीकरण `ax^(2)+bx+c =0` के मूल है तो `lim_(xto(1)/(alpha))sqrt((1-cos (cx^(2)+bx+a))/(2(1-alphax)^(2)))` का मान ज्ञात कीजिए। |
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Answer» प्रश्नानुसार `alpha ` व `beta ` समीकरण `ax^(2)+bx+c=0` के मूल है। `implies` समीकरण `cx^(2)+bx+a=0` के मूल `1/alpha`व `1/beta` होंगे `impliescx^(2)+bx+a=c(c-(1)/(alpha))(x-(1)/(beta))` `thereforelim_(xto(1)/(alpha))sqrt(1-cos(cx^(2)+bx+a))=lim_(xto(1)/(alpha))sqrt(1-cos {c(x-(1)/(alpha))(x-(1)/(beta))})` `=lim_(xto(1)/(alpha))|(sin {(c)/(2)(x-(1)/(alpha))(x-(1)/(beta))})/((1-alphax))|` `=lim_(xto(1)/(alpha))|(sin {(c)/(2)(x-(1)/(alpha))(x-(1)/(beta))})/(c/2(x-(1)/(alpha))(x-(1)/(beta)))*(c (alphax-1)(betax-1))/(2 alphabeta(1-alpha x))|` `=|(c)/(2 alpha beta)((beta)/(alpha)-1)|=|(c)/(2alpha)((1)/(alpha)-(1)/(beta))|` |
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| 22. |
परिगणना कीजिए `:lim_(xto(pi)/(2))(sqrt2-sqrt(1+sinx))/(sqrt2*cos ^(2)x)` |
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Answer» प्रश्नानुसार, `lim_(xto(pi)/(2))(sqrt2-sqrt(1+sinx))/(sqrt2*cos^(2)x)=lim_(xto(pi)/(2))(sqrt2-sqrt(1+sinx))/(sqrt2*cos^(2)x)xx(sqrt2+sqrt(1+sinx))/(sqrt2+sqrt(1+sinx))` `=lim_(xto(pi)/(2)){(2-(1+sinx))/(sqrt2(1-sin^(2)x))xx(1)/(sqrt2+sqrt(1+sinx))}` `=lim_(xto(pi)/(2)){((1-sinx))/(sqrt2(1-sinx)(1+sinx))xx(1)/(sqrt2+sqrt(1+sinx))}` `=lim_(xto(pi)/(2))(1)/(sqrt2(1+sinx))xx(1)/(sqrt2+sqrt(1+sinx))` `=(1)/(2sqrt2)xx(1)/(sqrt2xxsqrt2)` `=1/8` |
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| 23. |
मान ज्ञात कीजिए- `lim_(xto pi)(1-sin""(x)/(2))/(cos ""(x)/(2)(cos""(x)/(4)-sin""(x)/(4)))` |
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Answer» Correct Answer - `cos y` `lim_(xto pi)(1-sin x//2)/(x/2(cos ""(x)/(2)-sin""(x)/(4)))=lim_(xto pi)((1-sin x//2))/(cos ""(x)/(2)(cos ""(x)/(4)-sin""(x)/(4)))* (cos ""(x)/(4)+sin ""(x)/(4))/(cos ""(x)/(4)+sin ""(x)/(4))*(1+sin x//2)/(1+sin x//2)` `=lim_(xto pi)((1-sin^(2)x//2)(cos ""(x)/(4)+sin ""(x)/(4)))/(cos ""(x)/(2)(cos ^(2)""(x)/(4)-sin ^(2)""(x)/(4))(1+sin""(x)/(2)))` `=lim_(xto pi)((sin ""(x)/(4)+cos ""(x)/(4)))/(1+sin x//2)=((1)/(sqrt2)+(1)/(sqrt2))/(2)=(1)/(sqrt2)` |
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| 24. |
`lim_(xto pi//2)(cos x-cosec x)/((pi-2x)^(3))` का मान है-A. `1/16`B. `1/8`C. `1/4`D. `pi/2` |
| Answer» Correct Answer - A | |
| 25. |
`lim_(n tooo)(n!)/((n+1)!-n!)` का मान ज्ञात कीजिए। |
| Answer» `lim_(ntooo)(n!)/((n+1)!-n!)=lim_(ntooo)((n!)/((n+1)!))/(1-(n!)/((m+1)!))=lim_(n to oo)((1)/((n+1)))/(1-(1)/((n+1)))=(0)/(1-0)=(0)/(1-0)=0` | |
| 26. |
`lim_(xto pi//4)(int_(2)^(sec^(2)x)f(t)dt)/(x^(2)-(pi^(2))/(16))` का मान है-A. `8/pif(2)`B. `2/pif (2)`C. `2/pif((1)/(2))`D. `4f (2)` |
| Answer» Correct Answer - A | |
| 27. |
यदि `lim_(xto1)(x^(4)-1)/(x-1)=lim_(xtok)(x^(3)-k^(3))/(x^(2)-k^(2))` हो, तो k का मान होगा-A. `2/3`B. `4/3`C. `8/3`D. इनमे से कोई नहीं |
| Answer» Correct Answer - C | |
| 28. |
यदि `f(x)=(ax^(2)+b)/(x^(2)+1), lim_(xto0)f(x)=1` तथा `lim_(xtooo)f(x)=1` तब सिद्ध कीजिए कि `f(-2)=f(2)=1` |
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Answer» प्रश्नानुसार `f(x)=(ax^(2)+b)/(x^(2)+1)" "...(i)` तथा `lim_(xto0)f(x)=1implieslim_(xto0)(ax^(2)+b)/(x^(2)+1)=1impliesb=1` व `lim_(xtooo)f(x)=1implieslim_(xtooo)(ax^(2)+b)/(x^(2)+1)=1implieslim_(xtooo)(a+(b)/(x^(2)))/(1+(1)/(x^(2)))=1impliesa=1` अब, a व b के ये मान समीकरण (i ) में रखने पर `f(x)=(x^(2)+1)/(x^(2)+1)=1impliesf(2)=f(-2)=1` |
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| 29. |
`lim_(thetato(pi)/(2)) (sec theta-tan theta)` का मान ज्ञात कीजिये। |
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Answer» `underset(thetato(pi)/(2))lim (sec theta-tan theta)=underset(thetato (pi)/(2))lim((1)/(cos theta)-(sin theta)/(cos theta))=underset(x to (pi)/(2))lim((1-sin theta)/(cos theta))` `=underset(thetato (pi)/(2))lim(((1-sintheta)cos theta)/(cos^(2)theta))=underset(thetato (pi)/(2))lim((1-sintheta)cos theta)/(1-sin^(2)theta)` `=underset(thetato (pi)/(2))lim(cos theta)/(1+sin theta)=(0)/(1+1)=0` |
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| 30. |
`lim_(xto1) ((2x-3)(sqrtx-1))/(3x^(2)+3x-6)` का मान गये कीजिये। |
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Answer» `underset(x to 1)lim((2x-3)(sqrtx-1))/(3x^(2)+3x-6)=underset(xto1)lim((2x-3)(sqrtx-1)(sqrtx+1))/(3(x^(2)+x-2)(sqrtx+1))` `=underset(xto1)lim((2x-3))/(3(x+2)(x-1))((x-1)/(sqrtx+1))` `=underset(xto1)lim ((2x-3))/(3(x+2)(sqrtx+1))=(-1)/(18)` |
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| 31. |
`lim_(xtoa) (x^(m)-a^(m))/(x-a)` का मान ज्ञात कीजिये। |
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Answer» `underset(xtoa)lim(x^(m)-a^(m))/(x-a)` मान `x=a+h` यदि `x=a,` तब `h=0" "therefore x to a implieshto 0` `therefore underset(xtoa)lim (x^(m)-a^(m))/(x-a)=underset(hto0)lim((a+h)^(m)-a^(m))/(a+h-a)` `=underset(hto0)lim(a^(m)(1+(h)/(a))^(m)-a^(m))/(h)` `=underset(hto0)lim(a^(m))/(h)[(1+(h)/(a))^(m)-1]` `=underset(hto0)lim (a^(m))/(h)[1+m((h)/(a))+(m(m-1))/(2!)((h)/(a))+...-1]` (द्विपद प्रमेय से) `=underset(hto0)lim (a^(m))/(h)[m((h)/(a))+(m(m-1))/(2!)((h)/(a))+...]` `=underset(xto0)lim(a^(m))/(h)m((h)/(a))[1+((m-1))/(2!)((h)/(a))+...]` `=underset(hto0)lim m a^(m-1)[1+((m-1))/(2!)((h)/(a))+...]` `=m*a^(m-1)[1+0+0+...]` (सीमा लेने पर) `=ma^(m-1)` |
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| 32. |
`lim_(xtooo)((2x-3)(3x-4))/((4x-5)(5x-6))` का मान है-A. `1/10`B. 0C. `1/5`D. `3/10` |
| Answer» Correct Answer - D | |
| 33. |
`lim_(xto0)((2a)^(x)-(3b)^(x))/(x)`का मान होगा-A. `log ab `B. `log (2a//3b)`C. `log (b//a)`D. उपरोक्त में से कोई नहीं |
| Answer» Correct Answer - B | |
| 34. |
`lim_(xto0)(2sin^(2)3x)/(x^(2))` का मान होगा-A. 9B. 2C. 18D. 3 |
| Answer» Correct Answer - C | |
| 35. |
`lim_(xto0)(cosec x)^(1//logx )`का मान है- |
| Answer» Correct Answer - C | |
| 36. |
`lim_(xto0)(e^(x)+log (1+x)-(1-x)^(-2))/(x^(2))` का मान है- |
| Answer» Correct Answer - B | |
| 37. |
`lim_(x to pi//4)(sin x- cos x)/(x- pi//4)` का मान ज्ञात कीजिये। |
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Answer» `underset(x to pi//4)lim(sinx-cos x)/(x-pi//4)` माना `x=pi//4+h,` यदि `x=pi//4 impliesh=0` `therefore x to pi//4 impliesh to 0` `thereforeunderset(xto pi//4)lim(sin x-cos x)/(x- pi//4)=underset(xto0)lim(sin (pi//4+h)-cos (pi//4+h))/((pi)/(4)+h-(pi)/(h))` `=underset(hto0)lim((sinpi//4cos h+cos pi//4sin h)-(cos pi//4cos h-sin pi//4sin h))/(h)` `=underset(xto0)lim (1)/(sqrt2)((cos h+sin h-cos h+sin h))/(h)(cos pi//4=sin pi//4+(1)/(sqrt2))` `=(1)/(sqrt2)underset(hto0)lim (2 sin h)/(h)=sqrt2underset(hto0)lim(sin h)/(h)` `=sqrt2xx1=sqrt2" "(becauseunderset(h to 0)lim(sin h)/(h)=1)` |
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| 38. |
`lim_(xtooo) (x^(2)+bx+4)/(x^(2)+ax+5)` का मान है-A. `b/a`B. 0C. 1D. `4/5` |
| Answer» Correct Answer - C | |
| 39. |
`lim_(xto0) ((e^(x)-e^(-x))/(x))` का मान ज्ञात कीजिये। |
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Answer» हम जानते है कि `e^(x)=1+x+(x^(2))/(2!)+(x^(3))/(3!)+(x^(4))/(4!)+...oo` तथा `e^(-x)=1-x+(x^(2))/(2!)-(x^(3))/(3!)+(x^(4))/(4!)...oo` `therefore e^(x)-e^(-x)=2(x+(x^(3))/(3!)+(x^(5))/(5!)+...oo)` `=2x(1+(x^(2))/(3!)+(x^(4))/(5!)+...oo)` `therefore (e^(x)-e^(-x))/(x)=2[1+(x^(2))/(3!)+(x^(4))/(5!)+...oo]` अतः `underset(xto0)lim(e^(x)-e^(-x))/(x)=underset(xto0)lim2[1+(x^(2))/(3!)+(x^(4))/(5!)+...oo]` `=2[1+0+0+...]` (सिमा लेने पर ) `=2` |
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| 40. |
सिद्ध कीजिए कि `lim_(n to oo) n/2r^(2)sin ""(2pi)/(n)=pir^(2)` |
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Answer» `underset(nto oo)limn/2r^(2)sin ""(2pi)/(n)` यदि `n=oo` तब फलन का रूप `ooxx0` हो जाता है। `thereforeunderset(xto oo)limn/2r^(2)sin ""(2pi)/(n)=underset(x to oo)lim (r^(2)sin ""(2pi)/(n))/(2//n)` `=underset(x to oo)lim (pi r^(2)sin""(2pi)/(n))/((2pi)/(n))=pir^(2)underset( ntooo)lim (sin ""(2pi)/(n))/((2pi)/(n))` माना `(2pi)/(n)=theta` यदि `n=oo,` तब `theta=0` `therefore n to ooimpliesthetato 0` `therefore pir^(2)underset(n to oo)lim(sin (2pi//n))/(2pi//n)=pir^(2)underset(theta to 0)lim (sin theta)/(theta)` `=pi r^(2)xx` `=pir^(2)" "(इतिसिद्धम) " "(becauseunderset(thetato 0)lim(sin theta)/(theta)=1)` |
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| 41. |
`lim_(xtooo) x sin"" 1/x` का मान ज्ञात कीजिये। |
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Answer» `underset(xtooo)lim x sin ""1/x=underset(x to oo)lim (sin (1//2))/(1//x)` माना `1/x=theta` अतः यदि `x=oo` तब `theta=0` `therefore x to ooimpliestheta to 0` `therefore underset(xtooo)lim (sin 1//x)/((1//x))=underset(xto0)lim(sin theta)/(theta)=1` |
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