JEE Main Maths Limits, Continuity and Differentiability Previous Year Questions With Solutions

In calculus, limit and continuity are important concepts. A limit is a number that a function approaches as the independent variable of the function approaches a given value. A function can either be continuous or discontinuous. Positive Discontinuity, Jump Discontinuity and Infinite Discontinuity are types of discontinuity.  This article covers the definition of limit, types of limit, indeterminate form, algebra of limit, standard limits, expansion of some functions, continuity at a point, continuity in a domain, differentiability at a point. The limits, continuity and differentiability questions from the previous years of JEE Main are present on this page, along with the detailed solution for each question. These questions include all the important topics and formulae. About 2-3 questions are asked from this topic in JEE Examination.

JEE Main Maths Limits, Continuity and Differentiability Previous Year Questions With Solutions

Question 1: Solve limx1(2x3)(x1)2x2+x3\underset{x\to 1}{\mathop{\lim}}\,\frac{(2x-3)(\sqrt{x}-1)}{2{{x}^{2}}+x-3}

Solution:

limx1(2x3)(x1)×(x+1)(x1)(2x+3)×(x+1)=15.2=110\underset{x\to 1}{\mathop{\lim }}\,\,\,\frac{(2x-3)\,(\sqrt{x}-1)\times (\sqrt{x}+1)}{(x-1)\,(2x+3)\times (\sqrt{x}+1)}\\=\frac{-1}{5\,.\,2}\\=\frac{-1}{10}

Question 2: If f(9)=9, f'(9)=4, then limx9f(x)3x3\lim_{x \rightarrow 9}\frac{\sqrt{f(x)}-3}{\sqrt{x}-3}

Solution:

Applying L – Hospitals rule,

limx912f(x)f(x)12x=f(9)f(9)19=4313=4\underset{x\to 9}{\mathop{\lim }}\,\frac{\frac{1}{2\sqrt{f(x)}}\cdot {f}'(x)}{\frac{1}{2\sqrt{x}}}\\=\frac{\frac{{f}'(9)}{\sqrt{f(9)}}}{\frac{1}{\sqrt{9}}}\\=\frac{\frac{4}{3}}{\frac{1}{3}}\\=4

Question 3: Solve limh0(a+h)2sin(a+h)a2sinah\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}

Solution:

Apply L-Hospitals rule,

limh0(a+h)2sin(a+h)a2sinah\underset{h\to 0}{\mathop{\lim }}\,\,\frac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h} limh02(a+h)sin(a+h)+(a+h)2cos(a+h)1\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{2\,(a+h)\,\sin \,(a+h)+{{(a+h)}^{2}}\cos \,(a+h)}{1} =2asina+a2cosa=2a\,\,\sin a+{{a}^{2}}\cos \,\,a

Question 4: Solve limxπ/42cosx1cotx1\underset{x\to \pi /4}{\mathop{\lim }}\,\frac{\sqrt{2}\cos x-1}{\cot x-1}

Solution:

limxπ/4(2secx)cosx(1+cotx)cotx[2sec2x]=limxπ/4sinx(1+cotx)(2+secx)=12(2)2+2=12\underset{x\to \pi /4}{\mathop{\lim }}\,\,\frac{(\sqrt{2}-\sec x)\,\cos x\,(1+\cot x)}{\cot x\,[2-{{\sec }^{2}}x]} \\=\underset{x\to \pi /4}{\mathop{\lim }}\,\frac{\sin x\,(1+\cot x)}{(\sqrt{2}+\sec x)}\\=\frac{\frac{1}{\sqrt{2}}(2)}{\sqrt{2}+\sqrt{2}}\\=\frac{1}{2}

Question 5: Solve limx0[xtan12x]\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{x}{{{\tan }^{-1}}2x} \right]

Solution:

Let tan12x=θx=12tanθ and as x0,θ0limx0xtan12x=limθ012tanθθ=12{{\tan }^{-1}}2x=\theta \\\,\Rightarrow x=\frac{1}{2}\tan \theta \ and \ as \ x\to 0,\,\,\theta \to 0 \\\Rightarrow \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x}{{{\tan }^{-1}}2x}\\=\underset{\theta \to 0}{\mathop{\lim }}\,\,\frac{\frac{1}{2}\tan \theta }{\theta }\\=\frac{1}{2}

Question 6: Solve limx012(1cos2x)x\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\frac{1}{2}(1-\cos 2x)}}{x}

Solution:

limx012(1cos2x)x=limx0sinxx\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sqrt{\tfrac{1}{2}(1-\cos 2x)}}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{|\,\,\sin x\,\,|}{x}

So, limx0+sinxx=1\underset{x\to 0+}{\mathop{\lim }}\,\,\frac{|\,\sin x\,|}{x}=1 and

limx0sinxx=1\underset{x\to 0-}{\mathop{\lim }}\,\,\frac{|\,\sin x\,|}{x}=-1

Hence, the limit doesn’t exist.

Question 7: Solve limx0{tan(π4+x)}1/x\underset{x\to 0}{\mathop{\lim }}\,{{\left\{ \tan \left( \frac{\pi }{4}+x \right) \right\}}^{1/x}}

Solution:

Given limit = =limx0(1+tanx1tanx)1/x=limx0{(1+tanx)1/tanx}(tanx)/x{(1tanx)1/tanx}(tanx)/x=ee1=e2.=\underset{x\to 0}{\mathop{\lim }}\,\,\,{{\left( \frac{1+\tan x}{1-\tan x} \right)}^{1/x}} \\=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{\{{{(1+\tan x)}^{1/\tan x}}\}}^{(\tan x)/x}}}{{{\{{{(1-\tan x)}^{1/\tan x}}\}}^{(\tan x)/x}}}\\=\frac{e}{{{e}^{-1}}}\\={{e}^{2}}.

Question 8: Solve limx0(1+5x21+3x2)1/x2\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{1/{{x}^{2}}}}

Solution:

limx0(1+5x21+3x2)1/x2=limx0[(1+5x2)1/5x2]5limx0[(1+3x2)1/3x2]3=e5e3=e2\underset{x\to 0}{\mathop{\lim }}\,\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{1/{{x}^{2}}}}\\=\frac{\underset{x\to 0}{\mathop{\lim }}\,\,\,{{\left[ {{(1+5{{x}^{2}})}^{1/5{{x}^{2}}}} \right]}^{5}}}{\underset{x\to 0}{\mathop{\lim }}\,\,\,{{\left[ {{(1+3{{x}^{2}})}^{1/3{{x}^{2}}}} \right]}^{3}}}\\=\frac{{{e}^{5}}}{{{e}^{3}}}\\={{e}^{2}} [limx0(1+x)1/x=e][\because \,\,\,\underset{x\to 0}{\mathop{\lim }}\,\,{{(1+x)}^{1/x}}=e]

Question 9: Solve limx0xtan2x2xtanx(1cos2x)2\underset{x\to 0}{\mathop{\lim }}\,\frac{x\tan 2x-2x\tan x}{{{(1-\cos 2x)}^{2}}}

Solution:

limx0xtan2x2xtanx(1cos2x)2=limx0x(tan2x2tanx)(2sin2x)2=limx014x(tan2x2tanx)sin4x=limx014x{(2x+13(2x)3+215(2x)5+)2(x+x33+215x5+)}x4(1x23!+x45!+.)4=14.(8323)=24=12.\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x\tan 2x-2x\tan x}{{{(1-\cos \,\,2x)}^{2}}}\\=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x(\tan \,\,2x-2\tan x)}{{{(2\,{{\sin }^{2}}x)}^{2}}}\\=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1}{4}\,\frac{x\,(\tan 2x-2\tan x)}{{{\sin }^{4}}x}\\=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{4}\frac{x\left\{ \left( 2x+\frac{1}{3}{{(2x)}^{3}}+\frac{2}{15}\,{{(2x)}^{5}}+… \right)-2\left( x+\frac{{{x}^{3}}}{3}+\frac{2}{15}{{x}^{5}}+… \right) \right\}}{{{x}^{4}}\,{{\left( 1-\frac{{{x}^{2}}}{3\,\,!}+\frac{{{x}^{4}}}{5\,\,!}+…. \right)}^{4}}}\\=\frac{1}{4}\,.\,\left( \frac{8}{3}-\frac{2}{3} \right)\\=\frac{2}{4}\\=\frac{1}{2}.

Question 10: The function f(x)=log(1+ax)log(1bx)xf(x)=\frac{\log (1+ax)-\log (1-bx)}{x} is not defined at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is

Solution:

Since limit of a function is a + b as x → 0, therefore to be continuous at a function, its value must be a + b at x = 0

⇒ f (0) = a + b

Question 11: Evaluate f(x)={x3+x216x+20(x2)2if x2 kif x=2f(x)=\begin{array}{cc} \{ & \begin{array}{cc} \frac{{{x}^{3}}+{{x}^{2}}-16x+20}{{{(x-2)}^{2}}} & \text{if}\ x\ne 2\\ \\ \ k & \text{if}\ x=2 \end{array} \end{array}

Solution:

For continuous limx2f(x)=f(2)=kk=limx2x3+x216x+20(x2)2=limx2(x24x+4)(x+5)(x2)2=7.\underset{x\to 2}{\mathop{\lim }}\,\,f(x)=f(2)=k \\\Rightarrow \,\,k=\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{3}}+{{x}^{2}}-16x+20}{{{(x-2)}^{2}}} \\=\underset{x\to 2}{\mathop{\lim }}\,\,\frac{({{x}^{2}}-4x+4)\,\,(x+5)}{{{(x-2)}^{2}}}\\=7 .

Question 12: f(x)={x24x+3x21for x1 2for x=1f(x)=\begin{array}{cc} \{ & \begin{array}{cc} \frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1} & \text{for}\ x\ne 1 \\ \ 2 & \text{for }x=1 \end{array} \end{array}, then find the condition for the function to be continuous or discontinuous.

Solution:

f(x)={x24x+3x21}f(x)=\left\{ \frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1} \right\},

for x = 1

f (1) = 2, f(1+)=limx1+x24x+3x21=limx1+(x3)(x+1)=1f(1)=limx1x24x+3x21=1f(1)f(1)f(1+)=\underset{x\to 1+}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1}\\=\underset{x\to 1+}{\mathop{\lim }}\,\,\frac{(x-3)}{(x+1)}\\=-1 \\f(1-)=\underset{x\to 1-}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1}\\=-1\,\,\\\Rightarrow \,\,f(1)\ne f(1-)

Hence, the function is discontinuous at x = 1.

Question 13: Which of the following functions have a finite number of points of discontinuity in R ([.] represents the greatest integer function)?

A) tanx

B) x[x]

C) |x| / x

D) sin[πx]

Solution:

f (x) = tanx is discontinuous when x = (2n + 1) π / 2, n ∈ Z

f (x) = x[x] is discontinuous when x = k, k ∈ Z

f (x) = sin [nπx] is discontinuous when nπx = k, k ∈ Z

Thus, all the above functions have an infinite number of points of discontinuity. But, if (x) = |x| / x is discontinuous when x = 0 only.

Question 14: The number of values of x ∈ [0, 2] at which f (x) = ∣x − [1 / 2]∣ + |x − 1|+ tanx is not differentiable is

A) 0

B) 1

C) 3

D) None of these

Solution:

∣x − [1 / 2]∣ is continuous everywhere but not differentiable at x = 1 / 2, |x − 1| is continuous everywhere, but not differentiable at x = 1 and tan x is continuous in [0, 2] except at x = π / 2. Hence, f (x) is not differentiable at x = 1 / 2, 1, π / 2.

Question 15: limxπ/2(secθtanθ)=\underset{x\to \pi /2}{\mathop{\lim }}\,(\sec \theta -\tan \theta )=

Solution:

limθπ/21sinθcosθ=limθπ/2(cosθ2sinθ2)2(cosθ2sinθ2)(cosθ2+sinθ2)=0\underset{\theta \to \pi /2}{\mathop{\lim }}\,\,\,\frac{1-\sin \theta }{\cos \theta }=\underset{\theta \to \pi /2}{\mathop{\lim }}\,\,\,\frac{{{\left( \cos \frac{\theta }{2}-\sin \frac{\theta }{2} \right)}^{2}}}{\left( \cos \frac{\theta }{2}-\sin \frac{\theta }{2} \right)\,\left( \cos \frac{\theta }{2}+\sin \frac{\theta }{2} \right)}=0