JEE Advanced Maths PYQ – Limits (L’Hospital’s Rule, 1^?, Expansion Method)

Q 1 :

Let $k \in ℝ$ . If $\lim_{x \to 0^{+}}$ $(\sin {(\sin k x) + \cos x + x)}^{\frac{2}{x}}$ $= e^{6},$ then the value of $k$ is [2024]

1
2
3
4

1

2

3

4

(2)

Let, $ℓ = \lim_{x \to 0^{+}}$ $(\sin {(\sin k x) + \cos x + x)}^{\frac{2}{x}}$ $= e^{6}$

Taking log on both sides,

$\Rightarrow \ln ℓ = \lim_{x \to 0^{+}} \frac{2}{x} (\sin (\sin k x) + \cos x + x - 1)$

$\Rightarrow \ln ℓ = \lim_{x \to 0^{+}} 2 (\frac{\sin (\sin k x)}{\sin k x} \cdot \frac{\sin k x}{k x} \cdot \frac{k x}{x} + 1 - (\frac{1 - \cos x}{x^{2}}) x)$

$\Rightarrow \ln ℓ = 2 (k + 1)$ $\Rightarrow ℓ = e^{2 (k + 1)} = e^{6}$

$k + 1 = 3 \Rightarrow k = 2$

Q 2 :

If $\lim_{x \to 0} {[1 + x \ln (1 + b^{2})]}^{1 / x} = 2 b \sin^{2} θ, b > 0$ and $θ \in (- π, π],$ then the value of $θ$ is [2011]

$\pm \frac{π}{4}$
$\pm \frac{π}{3}$
$\pm \frac{π}{6}$
$\pm \frac{π}{2}$

1

2

3

4

(4)

$\lim_{x \to 0} {[1 + x \ln (1 + b^{2})]}^{\frac{1}{x}} = 2 b \sin^{2} θ$

$\Rightarrow e^{\lim_{x \to 0} \frac{1}{x} \ln [1 + x \ln (1 + b^{2})]} = 2 b \sin^{2} θ$

$\Rightarrow e^{\lim_{x \to 0} \frac{\ln [1 + x \ln (1 + b^{2})]}{x \ln (1 + b^{2})} \times \ln (1 + b^{2})} = 2 b \sin^{2} θ$

$\Rightarrow e^{\ln (1 + b^{2})} = 2 b \sin^{2} θ$

$\Rightarrow 1 + b^{2} = 2 b \sin^{2} θ$ $\Rightarrow 2 \sin^{2} θ = b + \frac{1}{b}$

We know that $2 \sin^{2} θ \leq 2$ and $b + \frac{1}{b} \geq 2 for b > 0$

$∴$ $2 \sin^{2} θ = b + \frac{1}{b} = 2 \Rightarrow \sin^{2} θ = 1$

$∵$ $θ \in (- π, π]$ $∴$ $θ = \pm \frac{π}{2}$

Q 3 :

$\lim_{x \to \frac{π}{4}} \frac{\int_{2}^{\sec^{2} x} f (t) d t}{x^{2} - \frac{π^{2}}{16}}$ equals [2007]

$\frac{8}{π} f (2)$
$\frac{2}{π} f (2)$
$\frac{2}{π} f (\frac{1}{2})$
$4 f (2)$

1

2

3

4

(1)
$\lim_{x \to \frac{π}{4}} \frac{\int_{2}^{\sec^{2} x} f (t) d t}{x^{2} - \frac{π^{2}}{16}}$ $[\frac{0}{0} -form]$

$= \lim_{x \to \frac{π}{4}} \frac{\frac{d}{d x} [\int_{2}^{\sec^{2} x} f (t) d t]}{\frac{d}{d x} (x^{2} - \frac{π^{2}}{16})}$ (Using L'Hospital rule)

$= \lim_{x \to \frac{π}{4}} \frac{f (\sec^{2} x) \cdot 2 \sec^{2} x \tan x}{2 x}$

$[∵ \frac{d}{d x} [\int_{g (x)}^{h (x)} f (t) d t] = f (h (x)) h' (x) - f (g (x)) g' (x)]$

$= \frac{f (2) \times 2 \times 2 \times 1}{2 \times \frac{π}{4}} = \frac{8}{π} f (2)$

Q 4 :

The value of $\lim_{x \to 0} ({(\sin x)}^{1 / x} + {(\frac{1}{x})}^{\sin x}),$ where $x > 0$ is [2006]

0
- 1
1
2

1

2

3

4

(3)

$\lim_{x \to 0} [{(\sin x)}^{1 / x} + {(\frac{1}{x})}^{\sin x}]$

$= \lim_{x \to 0} {(\sin x)}^{1 / x} + \lim_{x \to 0} {(\frac{1}{x})}^{\sin x}$

$= 0 + e^{\lim_{x \to 0} \sin x \log (\frac{1}{x})}$ $(∵ | \sin x | < 1 when x \to 0)$

$= e^{\lim_{x \to 0} \frac{- \log x}{cosec x}}$ $= e^{\lim_{x \to 0} \frac{- 1 / x}{- cosec x \cot x}}$ (Using L'Hospital rule)

$= e^{\lim_{x \to 0} \frac{\sin x}{x} \cdot \tan x}$ $= e^{0} = 1$

Q 5 :

If $f (x)$ is differentiable and strictly increasing function, then the value of $\lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ is [2004]

1
0
- 1
2

1

2

3

4

(3)

Let $L = \lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ $[using L.H. Rule]$

$[∵ f' (a) > 0 as f being strictly increasing]$

$L = \lim_{x \to 0} \frac{f' (x^{2}) \cdot 2 x - f' (x)}{f' (x)}$

$= \lim_{x \to 0} \frac{f' (x^{2}) \cdot 2 x}{f' (x)} - 1$

$= 0 - 1$

$= - 1$

Q 6 :

$\lim_{h \to 0} \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)},$ given that $f' (2) = 6$ and $f' (1) = 4$ [2003]

does not exist
is equal to $- \frac{3}{2}$
is equal to $\frac{3}{2}$
is equal to 3

1

2

3

4

(4)

$\lim_{h \to 0} \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)}$ $[\frac{0}{0} -form]$

$= \lim_{h \to 0} \frac{f' (2 h + 2 + h^{2}) \cdot (2 + 2 h)}{f' (h - h^{2} + 1) \cdot (1 - 2 h)}$ (Using L'Hospital rule)

$= \frac{f' (2) \cdot 2}{f' (1) \cdot 1} = \frac{6 \times 2}{4 \times 1} = 3$

Q 7 :

Let $f : R \to R$ be such that $f (1) = 3$ and $f' (1) = 6$ . Then $\lim_{x \to 0} {(\frac{f (1 + x)}{f (1)})}^{1 / x}$ equals [2002]

$1$
$e^{1 / 2}$
$e^{2}$
$e^{3}$

1

2

3

4

(3)

Given $f : R \to R,$ $f (1) = 3$ and $f' (1) = 6$

Then, $\lim_{x \to 0} {[\frac{f (1 + x)}{f (1)}]}^{1 / x}$

$= e^{\lim_{x \to 0} \frac{1}{x} [\log f (1 + x) - \log f (1)]}$

$= e^{\lim_{x \to 0} \frac{\frac{1}{f (1 + x)} f' (1 + x)}{1}}$ (Using L'Hospital rule)

$= e^{\frac{f' (1)}{f (1)}} = e^{6 / 3} = e^{2}$

Q 8 :

The integer $n$ for which $\lim_{x \to 0} \frac{(\cos x - 1) (\cos x - e^{x})}{x^{n}}$ is a finite non-zero number is [2002]

1
2
3
4

1

2

3

4

(3)

$\lim_{x \to 0} \frac{(\cos x - 1) (\cos x - e^{x})}{x^{n}}$

$= \lim_{x \to 0} \frac{(1 - \cos x) (1 + \cos x) (e^{x} - \cos x)}{x^{n} (1 + \cos x)}$

$= \lim_{x \to 0} (\frac{\sin^{2} x}{x^{2}}) \cdot (\frac{e^{x} - \cos x}{x^{n - 2}}) \cdot (\frac{1}{1 + \cos x})$

$= 1^{2} \cdot \frac{1}{2} \lim_{x \to 0} \frac{e^{x} - \cos x}{x^{n - 2}}$

$= \frac{1}{2} \lim_{x \to 0} \frac{e^{x} + \sin x}{(n - 2) x^{n - 3}}$ (Using L'Hospital rule)

For this limit to be finite, $n - 3 = 0$

$∴$ $n = 3$

Q 9 :

If $β = \lim_{x \to 0} \frac{e^{x^{3}} - {(1 - x^{3})}^{1 / 3} + ({(1 - x^{2})}^{1 / 2} - 1) \sin x}{x \sin^{2} x},$ then the value of $6 β$ is ____________. [2022]

(5)

$β = \lim_{x \to 0} \frac{e^{x^{3}} - {(1 - x^{3})}^{1 / 3} + ({(1 - x^{2})}^{1 / 2} - 1) \sin x}{x \cdot \frac{\sin^{2} x}{x^{2}} \cdot x^{2}}$

Using expansion,

$β = \lim_{x \to 0} [\frac{(1 + x^{3} + \frac{x^{6}}{2!} + \dots) - (1 - \frac{x^{3}}{3} + \frac{1}{3} \cdot \frac{- 2}{3} \cdot \frac{1}{2} x^{6} + \dots) + (- \frac{1}{2} x^{2} + \frac{1}{2} \cdot \frac{- 1}{2} \cdot \frac{1}{2} x^{4} + \dots) (x - \frac{x^{3}}{3!} + \dots)}{x^{3}}]$

$β = \lim_{x \to 0} \frac{x^{3} (1 + \frac{1}{3} - \frac{1}{2})}{x^{3}}$ (Neglecting higher powers of $x$ )

So, $β = \frac{5}{6}$

$∴$ $6 β = 5$

Q 10 :

Let $α, β \in ℝ$ be such that $\lim_{x \to 0} \frac{x^{2} \sin (β x)}{α x - \sin x} = 1 .$ Then $6 (α + β)$ equals ______. [2016]

(7)

$\lim_{x \to 0} \frac{x^{2} \sin (β x)}{α x - \sin x} = 1$

$\Rightarrow \lim_{x \to 0} \frac{x^{3} β}{α x - \sin x} = 1$

$\Rightarrow \lim_{x \to 0} \frac{x^{3} β}{α x - (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots \infty)} = 1$

$\Rightarrow \lim_{x \to 0} \frac{x^{3} β}{(α - 1) x + \frac{x^{3}}{3!} - \frac{x^{5}}{5!} + \dots \infty} = 1$

It is possible when

$α - 1 = 0$ and $β = \frac{1}{3!}$

$\Rightarrow α = 1$ and $β = \frac{1}{6}$

$∴$ $6 (α + β) = 6 (1 + \frac{1}{6}) = 7$

Q 11 :

Let $α$ be a positive real number. Let $f : ℝ \to ℝ$ and $g : (α, \infty) \to ℝ$ be the functions defined by $f (x) = \sin (\frac{π x}{12})$ and $g$ $(x)$ $= \frac{2 \log_{e} (\sqrt{x} - \sqrt{α})}{\log_{e} (e^{\sqrt{x}} - e^{\sqrt{α}})} .$

Then the value of $\lim_{x \to α^{+}} f (g (x))$ is ___________. [2022]

(00.50)

$We have, g$ $(x)$ $= \frac{2 \log_{e} (\sqrt{x} - \sqrt{α})}{\log_{e} (e^{\sqrt{x}} - e^{\sqrt{α}})}$

Now, $\lim_{x \to α^{+}} g$ $(x)$ $= \lim_{x \to α^{+}} \frac{\frac{2}{\sqrt{x} - \sqrt{α}} (\frac{1}{2 \sqrt{x}})}{\frac{1}{e^{\sqrt{x}} - e^{\sqrt{α}}} (\frac{1}{2 \sqrt{x}} e^{\sqrt{x}})},$

${when x \to α^{+}, f (x) \to \frac{\infty}{\infty}}$

Apply L.H. Rule

$= \lim_{x \to α^{+}} \frac{e^{\sqrt{x}} - e^{\sqrt{α}}}{\sqrt{x} - \sqrt{α}} \cdot \frac{1}{e^{\sqrt{x}}} \cdot 2$

$= \lim_{x \to α^{+}} \frac{e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}}{\frac{1}{2 \sqrt{x}}} \cdot \frac{2}{e^{\sqrt{x}}} = 2$

$= \lim_{x \to α^{+}} f (g (x)) = f (\lim_{x \to α^{+}} g (x)) = \sin \frac{π}{6} = \frac{1}{2} = 00.50$

Q 12 :

Let $S$ be the set of all $(α, β) \in ℝ \times ℝ$ such that $\lim_{x \to \infty} \frac{\sin (x^{2}) {(\log_{e} x)}^{α} \sin (\frac{1}{x^{2}})}{x^{α β} (\log_{e} {(1 + x))}^{β}} = 0 .$

Then which of the following is (are) correct? [2024]

$(- 1, 3) \in S$
$(- 1, 1) \in S$
$(1, - 1) \in S$
$(1, - 2) \in S$

1

2

3

4

Select one or more options

(2, 3)

$Given, \lim_{x \to \infty} \frac{\sin (x^{2}) \sin (\frac{1}{x^{2}}) {(\ln x)}^{α}}{x^{α β} (\ln {(1 + x))}^{β}} = 0$

$= \lim_{x \to \infty} \frac{(\sin x^{2}) \sin (\frac{1}{x^{2}}) \frac{1}{x^{2}} {(\ln x)}^{α}}{(\frac{1}{x^{2}}) x^{α β} (\ln {(1 + x))}^{β}} = 0$

$= \lim_{x \to \infty} {(\frac{\ln x}{\ln (1 + x)})}^{β} \cdot \frac{{(\ln x)}^{α - β}}{x^{α β + 2}} = 0$

$= \lim_{x \to \infty} \frac{{(\ln x)}^{α - β}}{x^{α β + 2}} = 0$

$It is possible if α β + 2 > 0 α β > - 2$

Topic Question Set

Q 1 :

Let $k \in ℝ$ . If $\lim_{x \to 0^{+}}$ $(\sin {(\sin k x) + \cos x + x)}^{\frac{2}{x}}$ $= e^{6},$ then the value of $k$ is [2024]

Q 2 :

If $\lim_{x \to 0} {[1 + x \ln (1 + b^{2})]}^{1 / x} = 2 b \sin^{2} θ, b > 0$ and $θ \in (- π, π],$ then the value of $θ$ is [2011]

Q 3 :

$\lim_{x \to \frac{π}{4}} \frac{\int_{2}^{\sec^{2} x} f (t) d t}{x^{2} - \frac{π^{2}}{16}}$ equals [2007]

Q 4 :

The value of $\lim_{x \to 0} ({(\sin x)}^{1 / x} + {(\frac{1}{x})}^{\sin x}),$ where $x > 0$ is [2006]

Q 5 :

If $f (x)$ is differentiable and strictly increasing function, then the value of $\lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ is [2004]

Q 6 :

$\lim_{h \to 0} \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)},$ given that $f' (2) = 6$ and $f' (1) = 4$ [2003]

Q 7 :

Let $f : R \to R$ be such that $f (1) = 3$ and $f' (1) = 6$ . Then $\lim_{x \to 0} {(\frac{f (1 + x)}{f (1)})}^{1 / x}$ equals [2002]

Q 8 :

The integer $n$ for which $\lim_{x \to 0} \frac{(\cos x - 1) (\cos x - e^{x})}{x^{n}}$ is a finite non-zero number is [2002]

Q 9 :

If $β = \lim_{x \to 0} \frac{e^{x^{3}} - {(1 - x^{3})}^{1 / 3} + ({(1 - x^{2})}^{1 / 2} - 1) \sin x}{x \sin^{2} x},$ then the value of $6 β$ is ____________. [2022]

Q 10 :

Let $α, β \in ℝ$ be such that $\lim_{x \to 0} \frac{x^{2} \sin (β x)}{α x - \sin x} = 1 .$ Then $6 (α + β)$ equals ______. [2016]

Q 12 :

Let $S$ be the set of all $(α, β) \in ℝ \times ℝ$ such that $\lim_{x \to \infty} \frac{\sin (x^{2}) {(\log_{e} x)}^{α} \sin (\frac{1}{x^{2}})}{x^{α β} (\log_{e} {(1 + x))}^{β}} = 0 .$

Then which of the following is (are) correct? [2024]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : Let k∈ℝ. If limx→0+(sin(sinkx)+cosx+x)2x=e6, then the value of k is [2024]

Q 2 : If limx→0[1+xln(1+b2)]1/x=2bsin2θ, b>0 and θ∈(-π,π], then the value of θ is [2011]

Q 3 : limx→π4∫2sec2xf(t)dtx2-π216 equals [2007]

Q 4 : The value of limx→0((sinx)1/x+(1x)sinx), where x>0 is [2006]

Q 5 : If f(x) is differentiable and strictly increasing function, then the value of limx→0f(x2)-f(x)f(x)-f(0) is [2004]

Q 6 : limh→0f(2h+2+h2)-f(2)f(h-h2+1)-f(1), given that f'(2)=6 and f'(1)=4 [2003]

Q 7 : Let f:R→R be such that f(1)=3 and f'(1)=6. Then limx→0(f(1+x)f(1))1/x equals [2002]

Q 8 : The integer n for which limx→0(cosx-1)(cosx-ex)xn is a finite non-zero number is [2002]

Q 9 : If β=limx→0ex3-(1-x3)1/3+((1-x2)1/2-1)sinxxsin2x, then the value of 6β is ____________. [2022]

Q 10 : Let α,β∈ℝ be such that limx→0x2sin(βx)αx-sinx=1. Then 6(α+β) equals ______. [2016]

Q 11 : Let α be a positive real number. Let f:ℝ→ℝ and g:(α,∞)→ℝ be the functions defined by f(x)=sin(πx12) and g(x)=2loge(x-α)loge(ex-eα). Then the value of limx→α+f(g(x)) is ___________. [2022]

Q 12 : Let S be the set of all (α,β)∈ℝ×ℝ such that limx→∞sin(x2)(logex)αsin(1x2)xαβ(loge(1+x))β=0. Then which of the following is (are) correct? [2024]

Want Us to Email you About Special Offers & Updates?

Q 1 :

Let $k \in ℝ$ . If $\lim_{x \to 0^{+}}$ $(\sin {(\sin k x) + \cos x + x)}^{\frac{2}{x}}$ $= e^{6},$ then the value of $k$ is [2024]

Q 2 :

If $\lim_{x \to 0} {[1 + x \ln (1 + b^{2})]}^{1 / x} = 2 b \sin^{2} θ, b > 0$ and $θ \in (- π, π],$ then the value of $θ$ is [2011]

Q 3 :

$\lim_{x \to \frac{π}{4}} \frac{\int_{2}^{\sec^{2} x} f (t) d t}{x^{2} - \frac{π^{2}}{16}}$ equals [2007]

Q 4 :

The value of $\lim_{x \to 0} ({(\sin x)}^{1 / x} + {(\frac{1}{x})}^{\sin x}),$ where $x > 0$ is [2006]

Q 5 :

If $f (x)$ is differentiable and strictly increasing function, then the value of $\lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ is [2004]

Q 6 :

$\lim_{h \to 0} \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)},$ given that $f' (2) = 6$ and $f' (1) = 4$ [2003]

Q 7 :

Let $f : R \to R$ be such that $f (1) = 3$ and $f' (1) = 6$ . Then $\lim_{x \to 0} {(\frac{f (1 + x)}{f (1)})}^{1 / x}$ equals [2002]

Q 8 :

The integer $n$ for which $\lim_{x \to 0} \frac{(\cos x - 1) (\cos x - e^{x})}{x^{n}}$ is a finite non-zero number is [2002]

Q 9 :

If $β = \lim_{x \to 0} \frac{e^{x^{3}} - {(1 - x^{3})}^{1 / 3} + ({(1 - x^{2})}^{1 / 2} - 1) \sin x}{x \sin^{2} x},$ then the value of $6 β$ is ____________. [2022]

Q 10 :

Let $α, β \in ℝ$ be such that $\lim_{x \to 0} \frac{x^{2} \sin (β x)}{α x - \sin x} = 1 .$ Then $6 (α + β)$ equals ______. [2016]

Q 12 :

Let $S$ be the set of all $(α, β) \in ℝ \times ℝ$ such that $\lim_{x \to \infty} \frac{\sin (x^{2}) {(\log_{e} x)}^{α} \sin (\frac{1}{x^{2}})}{x^{α β} (\log_{e} {(1 + x))}^{β}} = 0 .$

Then which of the following is (are) correct? [2024]