JEE Main Limits and Continuity — practice questions
Explore 2 free Limits and Continuity questions in JEE Main Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
A limit describes nearby values, while continuity also depends on the function value at the point. Try substitution first. If it gives 0/0, simplify the expression or use an appropriate standard limit, then compare both sides when the definition changes. These are original revision examples, not previous-year exam questions.
Key ideas and when to use them
Continuity at an interior point a requires f(a) to be defined and $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$.
For a polynomial, substitution gives the limit. For a rational function, substitution works when the denominator at the point is nonzero.
$\lim_{x\to0}\sin x/x=1$ when x is measured in radians. For real constants a and nonzero b, $\lim_{x\to0}\sin(ax)/(bx)=a/b$.
When x differs from a, $(x^2-a^2)/(x-a)=x+a$. Cancelling the factor helps find the limit; it does not define the original expression at x = a.
Common mistakes
0/0 is an indeterminate form, not a limit value and not proof that the limit fails to exist.
A removable hole can have a finite limit even though the function is undefined there. Continuity needs a matching value at that point.
For a piecewise function, check the left and right expressions separately. For example, |x|/x tends to −1 from the left and 1 from the right at zero, so no choice of f(0) makes it continuous there.
Standard trigonometric limits use radians. Converting to degrees without adjusting the formula changes the result.
Three worked examples
Example 1: Evaluate $\lim_{x\to3}(x^2-9)/(x-3)$.
Direct substitution gives 0/0. Factor the numerator as $(x-3)(x+3)$.
For x near 3 but not equal to 3, cancel x − 3. The remaining expression is x + 3.
Now substitute x = 3 in the simplified expression: 3 + 3 = 6. The original fraction remains undefined at 3.
The limit is 6.
Example 2: Evaluate $\lim_{x\to0}\sin(5x)/(2x)$, with angles in radians.
Rewrite the expression as $\frac52\cdot\frac{\sin(5x)}{5x}$.
As x tends to zero, 5x also tends to zero. The second factor tends to 1 by the standard sine limit.
The limit is $5/2$.
Example 3: For $x\ne4$, let $f(x)=(x^2-16)/(x-4)$, and let f(4) = k. Which k makes f continuous at 4?
For x ≠ 4, factor and simplify the fraction to x + 4.
Both one-sided limits at 4 equal 8, so the two-sided limit is 8.
Continuity requires f(4) to equal that limit. Therefore set k = 8.
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