Qus : 2 1 $\lim _{{x}\rightarrow1}\frac{{x}^4-1}{x-1}=\lim _{{x}\rightarrow k}\frac{{x}^3-{k}^2}{{x}^2-{k}^2}=$, then find k
1 8/3 2 4/3 3 2/3 4 1 Go to Discussion Solution
Quick DL Method Solution
Given:
\[
\lim_{x \to 1} \frac{x^4 - 1}{x - 1}
= \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2}
\]
LHS using derivative:
\[
\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4
\]
RHS using DL logic:
\[
\lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2}
\approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2}
\]
Equating both sides:
\[
\frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3}
\]
\[
\boxed{k = \frac{8}{3}}
\]
Qus : 3 3 Let $f(x)=\frac{x^2-1}{|x|-1}$. Then the value of $lim_{x\to-1} f(x)$ is
1 -1 2 1 3 2 4
3 Go to Discussion
Solution Qus : 5 4 The value of the limit $$\lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x}$$ is
1 1 2 $${3!}^{1/3!}$$ 3 $${3!}^{1/4}$$ 4 $${4!}^{1/4}$$ Go to Discussion
Solution Qus : 6 4 Which of the following is NOT true?
1 $$\lim _{{x}\rightarrow\infty}\frac{x}{{e}^x}=0$$ 2 $$\lim _{{x}\rightarrow0+}\frac{1}{{xe}^{1/x}}=0$$ 3 $$\lim _{{x}\rightarrow0+}\frac{\sin x}{{1+2x}}=0$$ 4 $$\lim _{{x}\rightarrow0+}\frac{\cos x}{{1+2x}}=0$$ Go to Discussion Qus : 7 4 For $a\in R$ (the set of al real numbers), $a \ne 1$, $\lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)(na+b)\ldots(na+n)\rbrack}=\frac{1}{60}$ . Then one of the value of $a$ is
1 5 2 8 3 -15/2 4 -17/2 Go to Discussion Qus : 8 3 The value of ${{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x}$ is equal to
1 2 2 1 3 0 4 -1 Go to Discussion Solution
Evaluate:
$$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}$$
Step 1: Apply L'Hôpital's Rule (since it's 0/0):
First derivative:
$$\frac{e^x + e^{-x} - 2}{\sin x}$$
Still 0/0 → Apply L'Hôpital's Rule again:
$$\frac{e^x - e^{-x}}{\cos x}$$
Now,
$$\lim_{x \to 0} \frac{1 - 1}{1} = 0$$
Final Answer:
$$\boxed{0}$$
Qus : 9 1 $\lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5}$ equal to
1 $e^{5}$ 2 $e^{-5}$ 3 $e^{2}$ 4 $e^{-2}$ Go to Discussion
Solution Qus : 10 1 Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If $\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$, then f(2) is
1 0 2 4 3 - 8 4 - 4 Go to Discussion Solution Given it has extremum values at x=1 and x=2
⇒f′(1)=0 and f′(2)=0
Given f(x) is a fourth degree polynomial
Let $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$
Given
$\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$
$\lim _{{x}\rightarrow0}\lbrack1+\frac{a{x}^4+b{x}^3+c{x}^2+\mathrm{d}x+e}{{x}^2}\rbrack=3$
$\lim _{{x}\rightarrow0}\lbrack1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{{x}^2}\rbrack=3$
For limit to have finite value, value of 'd' and 'e' must be 0
⇒d=0 & e=0
Substituting x=0 in limit
⇒ c+1=3
⇒ c=2
$f^{\prime}(x)=4a{x}^3+3b{x}^2+2cx+d$
$x=1$ and $x=2$ are extreme values,
⇒$f^{\prime}(1)=0$ and $f^{\prime}(2)=0
⇒ $4a+3b+4=0$ and $32a+12b+8=0$
By solving these equations
we get, $a=\frac{1}{2}$ and $b=-2$
So,
$f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$
⇒$f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)$
⇒$f(2)=0$
Qus : 11 4 The value of $\lim_{x\to a} \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$
1 $\frac{2}{3}$ 2 $\frac{2}{\sqrt{3}}$ 3 $\frac{3\sqrt{3}}{\sqrt{2}}$ 4 $\frac{2}{3\sqrt{3}}$ Go to Discussion
Solution Qus : 12 4 1 1/2 2 -1/2 3 - 2 4 2 Go to Discussion Solution Function is the form of
therefore using by L'Hospital rule
=
Again apply L'Hospital Rule,
=
Putting x = 0, we get
Qus : 13 2 Find
1 1 2 limit does not exist 3 infinity 4 None of these Go to Discussion
Solution Qus : 14 3 If $f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)}$ is a real number then $\lim _{{x}\rightarrow0}\, f(x)$
1 2 2 3 3 $$\log _e2$$ 4 $$\log _e3$$ Go to Discussion
Solution [{"qus_id":"3388","year":"2018"},{"qus_id":"3909","year":"2019"},{"qus_id":"4306","year":"2017"},{"qus_id":"4315","year":"2017"},{"qus_id":"9449","year":"2020"},{"qus_id":"10665","year":"2021"},{"qus_id":"11130","year":"2022"},{"qus_id":"11136","year":"2022"},{"qus_id":"11509","year":"2023"},{"qus_id":"11523","year":"2023"},{"qus_id":"11550","year":"2023"},{"qus_id":"11624","year":"2024"},{"qus_id":"11656","year":"2024"},{"qus_id":"10196","year":"2015"}]
is equal to 
therefore using by L'Hospital rule
