450 + Interview Questions in LIMITS IN GENERAL AWARENESS Page 4 InterviewSolution

151.	Evaluate `lim_(x to 0) (sinx+log(1-x))/(x^(2)).`
Answer» `underset(xto0)lim(sinx+log(1-x))/(x^(2))" "`(0/0 from) `=underset(xto0)lim((x-(x^(3))/(3!)+x^(5)/(5!)-* * * )+(-x-(x^(2))/(2)-(x^(3))/(3)-* * *))/(x^(2))` `=underset(xto0)lim((-x^(2))/(2)-x^(3)((1)/(3!)+(1)/(3))-(x^(4))/(4)...)/(x^(2))=-1/2`

Discussion

152.	Evaluate `lim_(xto0) ((1+x)^(1//x)-e+(1)/(2)es)/(x^(2))`.
Answer» `=(1+x)^(1//x)=e^((1)/(x)log(1+x))=e^((1)/(x)(x-(x^(2))/(2)+(x^(3))/(3)-...))` `=e^(1-(x)/(2)+(x^(2))/(3)-...)=e.e^(-(x)/(2)+(x^(2))/(3)-...)` `=e[1+(-(x)/(2)+(x^(2))/(3)-...)+(1)/(2!)(-(x)/(2)+(x^(2))/(3)...)^(2)+...]` `=e[1-(x)/(2)+(11)/(24)x^(2)-...]` Hence, `underset(xto0)lim((1+x)^(1//x)-e+(1)/(2)ex)/(x^(2))=(11e)/(24)`

Discussion

153.	Evaluate `underset(xto0)lim(5sinx-7sin2x+3sin3x)/(x^(2)sinx).`
Answer» `underset(xto0)lim(5sinx-7sin2x+3sin3x)/(x^(2)xinx)` `=underset(xto0)lim(5(x-(x^(3))/(3!)+...)-7(2x-(2x)^(3)/(3!)+...)+3(3x-(3x)^(3)/(3!)+...))/(x^(2)(x-(x^(3))/(3!)+...))` `=underset(xto0)lim((-5x^(3))/(3!)+(56x^(3))/(3!)-(81x^(3))/(3!))/(x^(3)(1-(x^(2))/(3!)+...))` `=(-5+56-81)/(3!)` `=-5`

Discussion

154.	`lim_(x->oo)((x+c)/(x-c))^x= 4` then find c.A. `log_(10)2`B. `log_e 2`C. `2`D. none of these
Answer» Correct Answer - B `lim_(xto oo) ((x+c)/(x-c^x))^x=4` ` rArr lim_(x to oo) (1+(2c)/(x-c))^x=4` `rArr e^(lim_(xto oo)(2cx)/(x-c))=4rArr e^(2c)=2rArr c=log_e 2` .

Discussion

155.	`lim_(xrarr oo) (1-(4)/(x-1))^(3x-1)` is equal toA. `e^12`B. `e^-12`C. `e^4`D. `e^3`
Answer» Correct Answer - B We have to find `=lim_(xto oo)(1-(4)/(x-1))^(3x-1)` `=lim_(xto oo)(1+(-4)/(x-1))^(3x-1)=e^(lim_(xto0^+)(-4(3x-1))/(x-1)=e^-12)`

Discussion

156.	Evaluate `underset(xto0)lim((1+x)^(1//x)-e+(1)/(2)es)/(x^(2))`.A. `(11e)/(24)`B. `(-11e)/(24)`C. `(e)/(24)`D. none of these
Answer» Correct Answer - A We have, `lim_(xto0) ((1+x)^(1//x)-e+(ex)/(2))/(x^2)` `lim_(xto0) ((1)/(e^x)log(1+x)_(-e+(ex)/2))/(x)` `lim_(xto0) (e^(1-(x)/(2)+(x^2)/(3)-(x^3)/(4)+.....)-e+(ex)/(2))/(x)` `=elim_(xto0) (e^(-(x)/(2)+(x^2)/(3)-(x^3)/(4)+.....)-1+(x)/(2))/(x^2)` `=elim_(xto0)(1+((x)/(2)+(x^2)/(3)-(x^3)/(4)+....)+(1)/(2ᴉ)(-(x)/(2)+(x^2)/(3)-(x^3)/(4)+...)^2 +....-1+(x)/(2))/(x^2)` `=lim_(xto0) (x^2((1)/(2)+(1)/(2!)xx(1)/(4))+x^3(...)+.....)/x^2=e((1)/(3)+(1)/(8))=(11e)/(24)`

Discussion

157.	The value of `lim_(xrarr0) ((1+x)^(1//x)-e)/(x)` isA. 1B. `(e)/(2)`C. `-(e)/(2)`D. `(2)/(e)`
Answer» Correct Answer - C We have , `(1+x)^(1//x)=e(log(1+x))/(x)=e(1)/(x)(x-(x^2)/(2)+(x^3)/(3)-...)` `rArr (1+x)^(1//x)=e^(1-(x)/(2)+(x^3)/(3) -..)= e.e (-x)/(2)+(x^2)/(3).....` `lim_(xto0)((1+x)^(1//x)-e)/(x)lim_(xto0)(e.e^(-(x)/2)+(x^2)/(3)....._(-e))/(x)` `e lim_(xto0) ({e^(-x//2+x^2//3+....)-1})/((-(x)/(2)+(x^2)/(3)+....))xx((-(x)/(2)+(x^2)/(x)+....))/(x)=-(1)/(2)e`

Discussion

158.	`lim_(xrarr0) (1+x+x^2-e^x)/(x^2)` is equal toA. 1B. 0C. `1//2`D. none of these
Answer» Correct Answer - C `lim_(xto0) (1+x+x^2-e^x)/(x^2)` ` =lim_(xto0) ((1+x+x^2)-(1+(x^2)/(2!)+(x^3)/(3!)+......))/(x^2)` ` lim_(x to 0) (1-(1)/(2!))-(x)/(3!)-(x^2)/(4!)+..... ` `=(1-(1)/(2))=(1)/(2)`

Discussion

159.	Let `p=lim_(x->0^+)(1+tan^2 sqrt(x))^(1/(2x))` then log p is equal to`A. `1`B. `(1)/(2)`C. `(1)/(4)`D. `2`
Answer» Correct Answer - B We have, `P=lim_(x to 0^+)(1+tan^2sqrt(x))^((1)/(2x))` `rArr P=e^(lim_(xto0^+)`

Discussion

160.	If `lim_(xrarr oo) (1+(a)/(x)-(4)/(x^2))^(2x)=e^3` , the `a` is equal toA. `(2)/(3)`B. `(3)/(2)`C. `2`D. `(1)/(2)`
Answer» Correct Answer - B `lim_(xto oo) (1+(a)/(x)-(4)/(x^2))^2x=e^3` `rArr e^(lim_(xto oo) 2x((a)/(x)-(4)/(x^2)))=e^3` `rArr lim_(xto oo) (2a-(8)/(x))=3 rArr 2a=3 rArr a=(3)/(2)`

Discussion

161.	If `lim_(xrarr oo) (1+(a)/(x)+(b)/(x^2))^(2x)=e^2`, thenA. `a=1,b=2`B. `a=2,b=1`C. `a=1,b in R`D. `a=b=1`
Answer» Correct Answer - C We have, `lim_(xto oo) (1+(a)/(x)+(b)/(x^2))^2x` `=e^(x^(lim_(x to oo)((a)/(x)+(b)/(2))xx2x))=e^xlim_(xto oo)(2a+(2b)/(x))=e^2a` `lim_(xto oo) (1+(a)/(x)+(b)/(x^2))^2x=e^2rArr e^2 rArr e^2a= e^2 rArr 2 rArr a=1`. Hence, `a=1 and b in R`.

Discussion

162.	If`("lim")_(xvec0)[1+x1n(1+b^2)]^(1/x)=2bsin^2theta,b >0,s mftheta in (-pi,pi],`then the value of `theta`is`+-pi/4`(b) `+-pi/3`(c) `+-pi/6`(d) `+-pi/2`A. `+-(pi)/(4)`B. `+-(pi)/(3)`C. `+-(pi)/(6)`D. `+-(pi)/(2)`
Answer» Correct Answer - D We have, `lim_(xto0) {1+x In (1+b^2)}^(1//x)=2bsin^2theta` `rArr e^(lim_(xto0)xIn (1+b^2)xx(1)/(x))=2bsin^2theta ` `rArr ^In (1+b^2=2bsin^2theta` `rArr 1+b^2=2bsin^2theta` `rArr sin^2theta (1+b^2)/(2b)` `rArr sin^2theta=(1)/(2)(b+(1)/(b))[because b+(1)/(b)ge 2therefore(1)/(2))ge 1]` `rArr sin^2theta=1` `rArr theta=+-(pi)/(2)`.

Discussion

163.	The graph of the function `y = f(x)` has a unique tangent at the point `(e^a,0)` through which the graph passes then `lim__(x->e^a) (log_e {1+7f (x)} - sin f(x))/(3f(x))`A. 1B. 2C. 7D. `-2`
Answer» Correct Answer - B

Discussion

164.	Using product rule,differentiate the following with respect to x:(i) y = x3sinx(ii) y = (x - 2)(x + 3)(iii) y = sinxcosx
Answer» (i) dy/dx = x³ d/dx(sinx) + sinx . d/dx x³ = x³cosx + sinx . 3x² = x³cosx + 3x²sinx (ii) dy/dx = (x - 2)d/dx(x + 3) (x + 3) d/dx(x - 2) = (x - 2)1 + (x + 3)1 = 2x + 1 (iii) dy/dx = sinx d/dx cosx + cosx d/dx sinx = snx . (-sinx) + cosxcosx = cos²x - sin²x

164.

Using product rule,differentiate the following with respect to x:(i) y = x3sinx(ii) y = (x - 2)(x + 3)(iii) y = sinxcosx

Answer»

(i) dy/dx = x³ d/dx(sinx) + sinx . d/dx x³

= x³cosx + sinx . 3x²

= x³cosx + 3x²sinx

(ii) dy/dx = (x - 2)d/dx(x + 3) (x + 3) d/dx(x - 2)

= (x - 2)1 + (x + 3)1

= 2x + 1

(iii) dy/dx = sinx d/dx cosx + cosx d/dx sinx

= snx . (-sinx) + cosxcosx

= cos²x - sin²x

Discussion

165.	The value of `lim_(xrarr 0) (1+sinx)^2` , isA. eB. `e^2`C. `sqrt(e)`D. none of these
Answer» Correct Answer - B We have, `lim_(xto0) (1+sinx)^(2cot x)=e^(lim_(xto1)sinx xx2cotx )=e^(lim_(xto0)2cosx)=e^2`

Discussion

166.	`lim_(x->1) (log_3 3x)^(log_x 3)=`A. eB. `(1)/(e)`C. `1`D. `-(1)/(e)`
Answer» Correct Answer - A We have, ` lim_(xto1) (log _3 3x)^log) x3` ` =lim_(xto1)(log _(3)3+log_(3)x)^log)x3` `lim_(xto1) (1+log_(3)x)(1)/(log_3x)=e ^(lim_(xto1))log _(3xx -(1)/(log_(3)x)=e^1=e`.

Discussion

167.	The value of `lim_(xrarr0) {(a^x+b^x+c^x)/(3)}^(1//x)`, isA. abcB. `(abc)^1//3`C. `(1)/(3)abc `D. none of these
Answer» Correct Answer - B `lim_(xto0) {(a^x+b^x+c^x)/(3)}^(1/x)` `lim_(xto0) {1+(a^x+b^x+c^x-3)/(3)}^(1/x)` `lim_(xto0) {1+((a^x-1)+(b^x-1)+(c^x-1))/(3)}^(1/x)` `e^(lim_(xto0) (a^x-1)/(3x)+(b^x-1)/(3x)+(c^x-1)/(3x)` `=e^((1)/(3){lim_(xto0) (a^x-1)/(3x)+lim_(xto0)(b^x-1)/(3x)+lim_(xto0)(c^x-1)/(3x)}` `=e^((1)/(3){log a+logb + log c}) =e^(log(abc) ^(1//3))=(abc) ^(1//3)`.

Discussion

168.	The value of `lim_(xrarr0)(cosx)^(cotx)`, isA. eB. `(1)/(e)`C. `1`D. `-1`
Answer» Correct Answer - C We have, `lim_(xto0) (cosx)^(cot x) =lim_(xto0) (1+cos x-1)^cot x`. ` =lim_(xto0) {1-2sin^2((x)/(2))}^(cotx )^=e^(lim_(xto0) -2sin^2(x//2).cotx` `=e^(lim_(x to 0) -2 )(sin^2(x//2)cosx)/(2sin(x//2)cosx//2)=e ^(lim_(xto0) -tan(x//2).cos x )=e^0=1`.

Discussion

169.	Evaluate : \(\lim\limits_{x \to 0}\frac{e^x-e^{sinx}}{x-sinx}\)
Answer» \(\lim\limits_{x \to 0}\frac{e^x-e^{sinx}}{x-sinx}\) \(=\lim\limits_{x \to 0}e^{sin\,x}(\frac{e^{x-sinx}-1}{x-sinx})\) \(=\lim\limits_{x \to 0}e^{sin\,x}.\lim\limits_{x \to 0}(\frac{e^{x-sinx}-1}{x-sinx})\) \(=e^{sin0}.1\) = 1

169.

Evaluate : \(\lim\limits_{x \to 0}\frac{e^x-e^{sinx}}{x-sinx}\)

Answer»

\(\lim\limits_{x \to 0}\frac{e^x-e^{sinx}}{x-sinx}\)

\(=\lim\limits_{x \to 0}e^{sin\,x}(\frac{e^{x-sinx}-1}{x-sinx})\)

\(=\lim\limits_{x \to 0}e^{sin\,x}.\lim\limits_{x \to 0}(\frac{e^{x-sinx}-1}{x-sinx})\)

\(=e^{sin0}.1\)

= 1

Discussion

170.	Evaluate : \(\lim\limits_{h \to 0}\frac{(a+h)^2sin(a+h)-a^2sin\,a}{h}\)
Answer» \( \lim\limits_{h \to 0}\frac{(a+h)^2sin(a+h)-a^2sin\,a}{h}\) \( \lim\limits_{h \to 0}\frac{(a^2+h^2+2ah)[sin\,a\,cosh+cosasinh)-a^2sin\,a}{h}\) \( \lim\limits_{h \to 0}[\frac{a^2\,sin\,a(cosh-1)}{h}\) \(\frac{a^2\,sin\,a\,sin\,h}{h}+(h+2a)(sin\,a\,cosh+cos\,a\,sin\,h)]\) = \( \lim\limits_{h \to 0}[\frac{a^2\,sin\,a(-2sin^2\frac{h}{2})}{h}.\frac{h}{2}]\) \(+ \lim\limits_{h \to 0}\frac{a^2\,cos\,a\,sin\,h}{h}+\lim\limits_{h \to 0}(h+2a)sin(a+h)\) = \(-a^2\,sin\,a.1^2.0+a^2\,cos\,a.1+2a\,sin\,a\) = \(a^2\,cosa+2a\,sin\,a.\)

170.

Evaluate : \(\lim\limits_{h \to 0}\frac{(a+h)^2sin(a+h)-a^2sin\,a}{h}\)

Answer»

\( \lim\limits_{h \to 0}\frac{(a+h)^2sin(a+h)-a^2sin\,a}{h}\)

\( \lim\limits_{h \to 0}\frac{(a^2+h^2+2ah)[sin\,a\,cosh+cosasinh)-a^2sin\,a}{h}\)

\( \lim\limits_{h \to 0}[\frac{a^2\,sin\,a(cosh-1)}{h}\)

\(\frac{a^2\,sin\,a\,sin\,h}{h}+(h+2a)(sin\,a\,cosh+cos\,a\,sin\,h)]\)

= \( \lim\limits_{h \to 0}[\frac{a^2\,sin\,a(-2sin^2\frac{h}{2})}{h}.\frac{h}{2}]\)

\(+ \lim\limits_{h \to 0}\frac{a^2\,cos\,a\,sin\,h}{h}+\lim\limits_{h \to 0}(h+2a)sin(a+h)\)

= \(-a^2\,sin\,a.1^2.0+a^2\,cos\,a.1+2a\,sin\,a\)

= \(a^2\,cosa+2a\,sin\,a.\)

Discussion

171.	`lim_(xrarr0)(2log(1+x)-log(1+2x))/(x^2)` is equal to
Answer» Correct Answer - B `lim_(xto0)(2log(1+x)-log(1+2x))/(x^2)` `=lim_(xto0) (log{(1+x)^2/(1+2x)})/(x^2)` `=lim_(xto0) log(1+(x^2)/(1+2x))/((x^2)/(1+2x))xx (1)/(1+2x)=1`

Discussion

172.	If `x_(1)` and `x_(2)` are the real and distinct roots of `ax^(2)+bx+c=0` then prove that `lim_(xtox1) (1+sin(ax^(2)+bx+c))^((1)/(x-x_(1)))=e^(a(x_(1)-x_(2))).`A. does not existB. 1C. `oo`D. `(1)/(2)`
Answer» `ax^(2)+bx+c=a(x-x_(1))(x-x_(2))` `underset(xtox1)lim(1+sin(ax^(2)+bx+c))^((1)/(x-x_(1)))" "(1^(oo))" form")` `=e^(underset(xtox_(1))lim(sin(a(x-x_(1))(x-x_(2))))/((x-x_(1))))` `=e^(underset(xtox_(1))lim(sin(a(x-x_(1)).(x-x_(2))))/(a(x-x_(1))(x-x_(2))).a(x-x_(2)))` `=e^(a(x_(1)-x_(2)))`

Discussion

173.	Evaluate `lim_(xto0) (log(5+x)-log(5-x))/(x).`
Answer» We have `underset(xto0)lim(log(5+x)-log(5-x))/(x)" "`(0/0 form) `=underset(xto0)lim(log{5(1+(x)/(5))}-log{5(1-(x)/(5))})/(x)` `=underset(xto0)lim({log5+log(1+(x)/(5))}-{log5+log(1- (x)/(5))})/(x)` `=underset(xto0)lim(log(1+(x)/(5))-log(1-(x)/(5)))/(x)` `=underset(xto0)lim(1)/(5)(log(1+(x)/(5))^(x))/(x//5)+underset(xto0)limlog(1-(x)/(5))/(-x//5)1/5=1/5+1/5=2/5`

Discussion

174.	Evaluate `lim_(hto0) (log_(e)(1+2h)-2log_(e)(1+h))/(h^(2)).`
Answer» `underset(hto0)lim(log_(e)(1+2h)-2log_(e)(1+h))/(h^(2))` `=underset(hto0)lim(log_(e)[(1+2h)/(1+2h+h^(2))])/(h^(2))` `=underset(hto0)lim(log_(e)[1+(-h^(2))/(1+2h+h^(2))])/((-h^(2))/(1+2h+h^(2)))xx(-1)/(1+2h+h^(2))` `=1xxunderset(hto0)lim(-1)/(1+2h+h^(2))` `=-1`

Discussion

175.	The value of ` lim_(hto0) (In (1+2h)-2ln (1+h))/(h^2)`, isA. 1B. -1C. 0D. none of these
Answer» Correct Answer - B

Discussion

176.	Evaluate `lim_(hto0) (2[sqrt(3)sin((pi)/(6)+h)-cos((pi)/(6)+h)])/(sqrt(3)h(sqrt(3)cosh-sinh)).`
Answer» Correct Answer - `4//3` `underset(hto0)lim(2[sqrt(3)sin((pi)/(6)+h)-cos((pi)/(6)+h)])/(sqrt(3)h(sqrt(3)cosh-sinh))` `underset(hto0)lim((4)/sqrt(3)[sqrt(3)/(2)sin((pi)/(6)+h)-(1)/(2)cos((pi)/(6)+h)])/(h(sqrt(3)cosh-sinh))` `=underset(hto0)lim(4)/(sqrt(3))xx(sinh)/(h)(1)/((sqrt(3)cosh-sinh))=4/3`

Discussion

177.	Lim x→∞((x/x+1)a + sin(1/x))x is equal to\(\lim\limits _{x\to\infty} \left(\left(\frac x {x+1}\right)^a+ sin\left(\frac{1}{x}\right)\right)^x\)
Answer» \(\lim\limits _{x\to\infty} \left(\left(\frac x {x+1}\right)^a+ sin\left(\frac{1}{x}\right)\right)^x\) \(= Exp\left[\lim\limits_{x \to\infty}\left(\left(\frac{x}{x +1}\right)^a + sin\left(\frac1x\right)-1\right) x\right]\) \(= Exp\left[\lim\limits_{x \to\infty} \left(\frac{x^{a+1} -(x +1)^a x}{(x +1)^a} + x\,sin\left(\frac 1 x\right)\right)\right]\) \(= Exp\left[\lim\limits_{x \to\infty}\left(\frac{x^{a+1}- (x^{a +1}+ ax^a + ....+x)}{x^a \left(1+\frac1x\right)^a}+x\left(\frac1x-\frac{1}{3!x^3}+...\right)\right)\right]\) \(= Exp\,[a+1]= e^{a+1}\) (By taking limit)

177.

Lim x→∞((x/x+1)a + sin(1/x))x is equal to\(\lim\limits _{x\to\infty} \left(\left(\frac x {x+1}\right)^a+ sin\left(\frac{1}{x}\right)\right)^x\)

Answer»

\(\lim\limits _{x\to\infty} \left(\left(\frac x {x+1}\right)^a+ sin\left(\frac{1}{x}\right)\right)^x\)

\(= Exp\left[\lim\limits_{x \to\infty}\left(\left(\frac{x}{x +1}\right)^a + sin\left(\frac1x\right)-1\right) x\right]\)

\(= Exp\left[\lim\limits_{x \to\infty} \left(\frac{x^{a+1} -(x +1)^a x}{(x +1)^a} + x\,sin\left(\frac 1 x\right)\right)\right]\)

\(= Exp\left[\lim\limits_{x \to\infty}\left(\frac{x^{a+1}- (x^{a +1}+ ax^a + ....+x)}{x^a \left(1+\frac1x\right)^a}+x\left(\frac1x-\frac{1}{3!x^3}+...\right)\right)\right]\)

\(= Exp\,[a+1]= e^{a+1}\) (By taking limit)

Discussion

178.	Solve: \(\lim\limits_{x \to 0}(1 + 3x)^{\frac{1}{x}}\)
Answer» = \(\lim\limits_{x \to 0}\Big[(1 + 3x)^{\frac{1}{3x}}\Big]^3\) = [e]³ = e³

Discussion

179.	Solve: \(\lim\limits_{n \to \infty}\Big(1 + \frac{2}{n}\Big)^n\)
Answer» \(= \lim\limits_{n \to \infty}\Big[\Big(1 + \frac{1}{\frac{n}{2}}\Big)^{\frac{n}{2}}\Big]^2\) = [e]² = e².

Discussion

180.	Solve: \(\lim\limits_{x \to 0}\frac{e^{-3x} - 1}{x}\)
Answer» = \(\lim\limits_{x \to 0}\frac{e^{-3x} - 1}{-3x} \times - 3\) = - 3

180.

Solve: \(\lim\limits_{x \to 0}\frac{e^{-3x} - 1}{x}\)

Answer»

= \(\lim\limits_{x \to 0}\frac{e^{-3x} - 1}{-3x} \times - 3\)

= - 3

Discussion

181.	Solve: \(\lim\limits_{x \to 0}\Big(\frac{2^x - 1}{3x}\Big)\)
Answer» = \(\lim\limits_{x \to 0}\frac{2^x - 1}{x} \times \frac{1}{3}\) = log_e2 x \(\frac{1}{3}\) = \(\frac{log_e2}{3}\)

181.

Solve: \(\lim\limits_{x \to 0}\Big(\frac{2^x - 1}{3x}\Big)\)

Answer»

= \(\lim\limits_{x \to 0}\frac{2^x - 1}{x} \times \frac{1}{3}\)

= log_e2 x \(\frac{1}{3}\) = \(\frac{log_e2}{3}\)

Discussion

182.	Evaluate :\(\lim\limits_{x \to 0}\frac{log(6+x)-log(6-x)}{x}\)
Answer» \(\lim\limits_{x \to 0}\frac{log(1+\frac{x}{6})-log6(1-\frac{x}{6})}{x}\) = \(\lim\limits_{x \to 0}\frac{[log6+log(1+\frac{x}{6}]-[log6+log(1-\frac{x}{6})]}{x}\) = \(\lim\limits_{x \to 0}[\frac{log(1+\frac{x}{6})}{x}-\frac{log(1-\frac{x}{6})}{x}]\) = \(\lim\limits_{x \to 0}.\frac{1}{6}\frac{log(1+\frac{x}{6})}{\frac{x}{6}}+\lim\limits_{x \to 0}.\frac{1}{6}\frac{log(1-\frac{x}{6})}{(-\frac{x}{6})}]\) = \(\frac{1}{6}\times1+\frac{1}{6}\times1\,[∵ \lim\limits_{x \to 0}\frac{log(1+x)}{x}=\lim\limits_{x \to 0}\frac{log(1-x)}{-x}=1]\) = 0

182.

Evaluate :\(\lim\limits_{x \to 0}\frac{log(6+x)-log(6-x)}{x}\)

Answer»

\(\lim\limits_{x \to 0}\frac{log(1+\frac{x}{6})-log6(1-\frac{x}{6})}{x}\)

= \(\lim\limits_{x \to 0}\frac{[log6+log(1+\frac{x}{6}]-[log6+log(1-\frac{x}{6})]}{x}\)

= \(\lim\limits_{x \to 0}[\frac{log(1+\frac{x}{6})}{x}-\frac{log(1-\frac{x}{6})}{x}]\)

= \(\lim\limits_{x \to 0}.\frac{1}{6}\frac{log(1+\frac{x}{6})}{\frac{x}{6}}+\lim\limits_{x \to 0}.\frac{1}{6}\frac{log(1-\frac{x}{6})}{(-\frac{x}{6})}]\)

= \(\frac{1}{6}\times1+\frac{1}{6}\times1\,[∵ \lim\limits_{x \to 0}\frac{log(1+x)}{x}=\lim\limits_{x \to 0}\frac{log(1-x)}{-x}=1]\)

= 0

Discussion

183.	Show that the function f(x) = \(\begin{cases} (1 + 3x)^{\frac{1}{x}} & x \neq 0 \\e^3 &x = 0 \end{cases}\) is continuous at x = 0
Answer» \(\lim\limits_{x \to 0}(1 + 3x)^{\frac{1}{3x}\times 3}\) = e³ Also f(x) at x = 0 is e³ i.e f(0) = e³ ∴ \(\lim\limits_{x \to 0}\) f(x) = f(0) = e³ ∴ f(x) is continuous at x = 0

183.

Show that the function f(x) = \(\begin{cases} (1 + 3x)^{\frac{1}{x}} & x \neq 0 \\e^3 &x = 0 \end{cases}\) is continuous at x = 0

Answer»

\(\lim\limits_{x \to 0}(1 + 3x)^{\frac{1}{3x}\times 3}\) = e³

Also f(x) at x = 0 is e³

i.e f(0) = e³

∴ \(\lim\limits_{x \to 0}\) f(x) = f(0) = e³

∴ f(x) is continuous at x = 0

Discussion

184.	Show f(x) defined by f(x) = \(\begin{cases}\frac{x^2 - 25}{x - 5} &\; when\; x \neq 5\\10& \;when\; x = 5\end{cases}\) is continuous at x = 5
Answer» Also f(x) at x = 5 is 10; i.e, f(5) = 10 Here \(\lim\limits_{x \to 5}\) f(x) = f(5) = 10 ∴ the function f(x) is continuous at x = 5

184.

Show f(x) defined by f(x) = \(\begin{cases}\frac{x^2 - 25}{x - 5} &\; when\; x \neq 5\\10& \;when\; x = 5\end{cases}\) is continuous at x = 5

Answer»

Also f(x) at x = 5 is 10; i.e, f(5) = 10

Here \(\lim\limits_{x \to 5}\) f(x) = f(5) = 10

∴ the function f(x) is continuous at x = 5

Discussion

185.	Find k for which f(x) = \(\begin{cases}k + x, & x = 1\\4x + 3, & x \neq 1\end{cases}\) is continuous at x = 1.
Answer» \(\lim\limits_{x \to 1}f(x) = \lim\limits_{x \to 1}\)(4x + 3) = 4 + 3 = 7 since f(x) is continuous at x = 1 \(\lim\limits_{x \to 1}f(x) \) = f(1) 7 = k + 1; k = 7 - 1 = 6

185.

Find k for which f(x) = \(\begin{cases}k + x, & x = 1\\4x + 3, & x \neq 1\end{cases}\) is continuous at x = 1.

Answer»

\(\lim\limits_{x \to 1}f(x) = \lim\limits_{x \to 1}\)(4x + 3)

= 4 + 3 = 7

since f(x) is continuous at x = 1

\(\lim\limits_{x \to 1}f(x) \) = f(1)

7 = k + 1;

k = 7 - 1 = 6

Discussion

186.	Find lim(x→1) (x3 - x2 + 1)
Answer» lim_(x→1) (x³ - x²+ 1) = 1³ - 1² + 1 = 1

Discussion

187.	Find lim(x → 1) (x2 + x)
Answer» lim_(x→1)(x² + x) = 1² + 1 = 2

Discussion

188.	Find lim(x→a) (x3 - a3)/(x - a)
Answer» lim_(x→a) (x³ - a³)/(x - a) = 3a²

Discussion

189.	Differentiate the following with respect to x(i) y = a/x4 - b/x2 + cosx(ii) y = sin(x + a)
Answer» (i) y = ax^-4 - bx^-2 + cosx ∴ dy/dx = a(-4x^-5) - b(-2x^-3) - sinx = - 4ax^-5 + 2bx^-3 - sinx (ii) y = sinxcosa + sinacosx ∴ dy/dx = cosa. dy/dx sinx + sina d/dxcosx = cosa . cosx - sinasinx = cos(x + a)

189.

Differentiate the following with respect to x(i) y = a/x4 - b/x2 + cosx(ii) y = sin(x + a)

Answer»

(i) y = ax^-4 - bx^-2 + cosx

∴ dy/dx = a(-4x^-5) - b(-2x^-3) - sinx

= - 4ax^-5 + 2bx^-3 - sinx

(ii) y = sinxcosa + sinacosx

∴ dy/dx = cosa. dy/dx sinx + sina d/dxcosx

= cosa . cosx - sinasinx

= cos(x + a)

Discussion

190.	Differentiate the following with respect to x(i) y = x + a(ii) y = 4√x - 2
Answer» (i) dy/dx = 1 + 0 = 1 (ii) ∴ dy/dx = 4(1/2x^-1/2) - 0 = 2x^-1/2

190.

Differentiate the following with respect to x(i) y = x + a(ii) y = 4√x - 2

Answer»

(i) dy/dx = 1 + 0 = 1

(ii) ∴ dy/dx = 4(1/2x^-1/2) - 0 = 2x^-1/2

Discussion

191.	If `a_(1)=1` and `a_(n)+1=(4+3a_(n))/(3+2a_(n)),nge1"and if" lim_(ntooo) a_(n)=a,"then find the value of a."`
Answer» We have `a_(n)+1=(4+3a_(n))/(3+2a_(n))` `:." "underset(ntooo)lima_(n)+1=underset(ntooo)lim(4+3an)/(3+2a_(n)` `implies" "underset(ntooo)lima_(n)+1=(4+3underset(ntooo)lima_(n))/(3+2underset(ntooo)lima_(n))` `a=(4+3a)/(3+2a)" "(becauseunderset(ntooo)lima_(n)+1=underset(ntooo)lima_(n)=a)` `2a^(2)=4` `a=sqrt(2)" "(ane-sqrt(2)"because"a_(n)gt0)`

Discussion

192.	Evaluate `lim_(xto0) ((a^(x)+b^(x)+c^(x))/(3))^(2//x),(a,b,cgt0)`
Answer» We have `underset(xto0)lim((a^(x)+b^(x)+c^(x))/(3))^(2//x)=e^(underset(xto0)lim((a^(x)+b^(x)+c^(x))/(3)-1).2/x` `=e^(2/3underset(xto0)lim((a^(x)+b^(x)+c^(x)-3)/(x))` `=e^(2/3underset(xto0)lim((a^(x)-1)/(x)+(b^(x)-1)/(x)+(c^(x)-1)/(x))` `=e^(2/3{underset(xto0)lima^(x-1)/(x)+underset(xto0)lim(b^(x)-1)/(x)+underset(xto0)lim(c^(x)-1)/(x)}}` `=e^((2//3){"In "a+"In "b+"In "c}}` `=e^((2//3)"In"(abc))` `=e^("In"(abc)^(2//3))` `=(abc)^(2//3)`

Discussion

193.	`lim_(x rarr 2) (x^(-8)-(1)/(256))/(x-2)=`______.A. `-(1)/(32)`B. `-(1)/(128)`C. `-(1)/(256)`D. `-(1)/(64)`
Answer» Correct Answer - D Use the formula, `underset(x rarr a)("lim")(x^(n)-a^(n))/(x-a)=na^(n-1)`.

Discussion

194.	`lim_(x rarr 2) (x^(2)+2x-8)/(2x^(2)-3x-2)=`______.A. `(1)/(5)`B. `(6)/(5)`C. `(3)/(2)`D. `-(1)/(6)`
Answer» Correct Answer - B Factorize the numerator and denominator and remove common factor. Now substitute x = 2.

Discussion

195.	` lim_(x rarr 4^(-)) sqrt(16-x^(2))` is a/anA. complex number.B. real number.C. natural number.D. integer.
Answer» Correct Answer - B Apply the concept of left hand limit.

Discussion

196.	`lim_(x rarr -3) (x^(2)-2x-15)/(x^(2)+2x-3)=`______.A. 1B. 2C. -3D. -4
Answer» Correct Answer - B Factorize the numerator and denominator and cancel the common factor and then substitute x = -3.

Discussion

197.	Evaluate: `lim_(x rarr a) [(x^(n)-a^(n))/(x^(m)-a^(m))]`.
Answer» Correct Answer - `(n)/(m)a^(n-m)`

Discussion

198.	`lim_(x rarr 0) log((2x+1)/(5x+4))=`______.A. `-2log 2`B. log 4C. `log((1)/(2))`D. `-3log 2`
Answer» Correct Answer - A Substitute x = 0.

Discussion

199.	`lim_(x rarr 5) (sqrt(x+20)-sqrt(3x+10))/(5-x)=`______.A. `-(2)/(5)`B. `(2)/(5)`C. `(1)/(5)`D. `(-1)/(5)`
Answer» Correct Answer - C Rationalize the numerator and cancel common factor. Then substitute, x = 5.

Discussion

200.	Evaluate: `lim_(x rarr 3) (\|x-3\|)/(x-3)`.
Answer» Correct Answer - Limit does not exist

Discussion

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Show f(x) defined by f(x) = \(\begin{cases}\frac{x^2 - 25}{x - 5} &amp;\; when\; x \neq 5\\10&amp; \;when\; x = 5\end{cases}\) is continuous at x = 5

Find k for which f(x) = \(\begin{cases}k + x, &amp; x = 1\\4x + 3, &amp; x \neq 1\end{cases}\) is continuous at x = 1.

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Differentiate the following with respect to x(i) y = a/x4 - b/x2 + cosx(ii) y = sin(x + a)

Differentiate the following with respect to x(i) y = x + a(ii) y = 4√x - 2

If `a_(1)=1` and `a_(n)+1=(4+3a_(n))/(3+2a_(n)),nge1"and if" lim_(ntooo) a_(n)=a,"then find the value of a."`

Evaluate `lim_(xto0) ((a^(x)+b^(x)+c^(x))/(3))^(2//x),(a,b,cgt0)`

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`lim_(x rarr 0) log((2x+1)/(5x+4))=`______.A. `-2log 2`B. log 4C. `log((1)/(2))`D. `-3log 2`

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Evaluate: `lim_(x rarr 3) (|x-3|)/(x-3)`.

Show that the function f(x) = \(\begin{cases} (1 + 3x)^{\frac{1}{x}} & x \neq 0 \\e^3 &x = 0 \end{cases}\) is continuous at x = 0

Show f(x) defined by f(x) = \(\begin{cases}\frac{x^2 - 25}{x - 5} &\; when\; x \neq 5\\10& \;when\; x = 5\end{cases}\) is continuous at x = 5

Find k for which f(x) = \(\begin{cases}k + x, & x = 1\\4x + 3, & x \neq 1\end{cases}\) is continuous at x = 1.