Evaluate lim (x→a) (√(a + 2x) - √(3x))...

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 1978Easy

Evaluate $lim_{x \to a} \frac{a + 2 x - 3 x}{3 a + x - 2 x}, (a \neq = 0) .$

Answer:

\frac{2}{3 3}

Visualized Solution (Hindi)

Checking the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \frac{0}{0} Form

Check the form of the limit as $x \to a$
Numerator: $a + 2 a - 3 a = 3 a - 3 a = 0$
Denominator: $3 a + a - 2 a = 4 a - 2 a = 2 a - 2 a = 0$
This is a $\frac{0}{0}$ indeterminate form.

Rationalizing the Numerator

Rationalize the numerator by multiplying by its conjugate: $a + 2 x + 3 x$
Numerator transformation: $(a + 2 x - 3 x) (a + 2 x + 3 x) = (a + 2 x) - 3 x = a - x$

Rationalizing the Denominator

Rationalize the denominator by multiplying by its conjugate: $3 a + x + 2 x$
Denominator transformation: $(3 a + x - 2 x) (3 a + x + 2 x) = (3 a + x) - 4 x = 3 a - 3 x = 3 (a - x)$

Simplifying the Expression

Combine the rationalized parts: $lim_{x \to a} \frac{( a - x ) ( 3 a + x + 2 x )}{3 ( a - x ) ( a + 2 x + 3 x )}$
Cancel the common factor $(a - x)$ since $x \neq = a$

Evaluation and Final Answer

Substitute $x = a$ into the simplified expression: $\frac{3 a + a + 2 a}{3 ( a + 2 a + 3 a )}$
Evaluate: $\frac{2 a + 2 a}{3 ( 3 a + 3 a )} = \frac{4 a}{3 ( 2 3 a )}$
Final simplification: $\frac{4 a}{6 3 a} = \frac{2}{3 3}$

Key Takeaways

Key Takeaway: Rationalization is the primary tool for limits involving square roots that result in $\frac{0}{0}$ forms.
Next Challenge: Try evaluating the same limit using L'Hopital's Rule to verify the result.

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Conceptually Similar Problems

MathematicsTechniques of DifferentiationJEE Advanced 1980Easy

View in:English Hindi

Evaluate: $lim_{h \to 0} \frac{( a + h ) ^{2} s i n ( a + h ) - a ^{2} s i n a}{h}$

Answer:

a^{2} cos a + 2 a sin a

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 2000Easy

View in:English Hindi

For $x \in R, lim_{x \to \infty} (\frac{x - 3}{x + 2})^{x} =$

(A)

e

(B)

e^{- 1}

(C)

e^{- 5}

(D)

e^{5}

Answer:C

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 1993Easy

View in:English Hindi

Find $lim_{x \to 0} {tan (π /4 + x)}^{1/ x} .$

Answer:

e^{2}

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 1996Easy

View in:English Hindi

$lim_{x \to 0} (\frac{1 + 5 x ^{2}}{1 + 3 x ^{2}})^{1/ x^{2}} = ...$

Answer:

e^{2}

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 1979Easy

View in:English Hindi

$f (x)$ is the integral of $\frac{2 s i n x - s i n 2 x}{x ^{3}}, x \neq = 0$ , find $lim_{x \to 0} f^{'} (x)$

Answer:1

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2002Easy

View in:English Hindi

$lim_{x \to \infty} (\frac{x ^{2} + 5 x + 3}{x ^{2} + x + 3})^{x}$

(A)

e^4

(B)

e^2

(C)

e^3

(D)

Answer:A

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2015Easy

View in:English Hindi

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

(A)

(B)

1/2

(C)

(D)

Answer:A

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 1987Easy

View in:English Hindi

$lim_{x \to - \infty} [\frac{x ^{4} s i n ( 1/ x ) + x ^{2}}{( 1 + ∣ x ∣ ^{3} )}] = .........$

Answer:

- 1

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 1979Easy

View in:English Hindi

If $f (x) = \frac{x - s i n x}{x + c o s ^{2} x}$ , then $lim_{x \to \infty} f (x)$ is

(A)

(B)

\infty

(C)

(D)

none of these

Answer:C

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2003Easy

View in:English Hindi

$lim_{x \to \frac{π}{2}} \frac{1 - t a n ( x /2 )}{1 + t a n ( x /2 )} \frac{[ 1 - s i n x ]}{[ π - 2 x ] ^{3}}$ is

(A)

\infty

(B)

1/8

(C)

(D)

1/32

Answer:D

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate limx→a​3a+x​−2x​a+2x​−3x​​,(a=0).

Visualized Solution (Hindi)

Checking the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 00​ Form

Rationalizing the Numerator

Rationalizing the Denominator

Simplifying the Expression

Evaluation and Final Answer

Key Takeaways

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Evaluate $lim_{x \to a} \frac{a + 2 x - 3 x}{3 a + x - 2 x}, (a \neq = 0) .$

Checking the $\frac{0}{0}$ Form

Evaluate: $lim_{h \to 0} \frac{( a + h ) ^{2} s i n ( a + h ) - a ^{2} s i n a}{h}$

For $x \in R, lim_{x \to \infty} (\frac{x - 3}{x + 2})^{x} =$

Find $lim_{x \to 0} {tan (π /4 + x)}^{1/ x} .$

$lim_{x \to 0} (\frac{1 + 5 x ^{2}}{1 + 3 x ^{2}})^{1/ x^{2}} = ...$

$f (x)$ is the integral of $\frac{2 s i n x - s i n 2 x}{x ^{3}}, x \neq = 0$ , find $lim_{x \to 0} f^{'} (x)$

$lim_{x \to \infty} (\frac{x ^{2} + 5 x + 3}{x ^{2} + x + 3})^{x}$

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to - \infty} [\frac{x ^{4} s i n ( 1/ x ) + x ^{2}}{( 1 + ∣ x ∣ ^{3} )}] = .........$

If $f (x) = \frac{x - s i n x}{x + c o s ^{2} x}$ , then $lim_{x \to \infty} f (x)$ is

$lim_{x \to \frac{π}{2}} \frac{1 - t a n ( x /2 )}{1 + t a n ( x /2 )} \frac{[ 1 - s i n x ]}{[ π - 2 x ] ^{3}}$ is

Evaluate $lim_{x \to a} \frac{a + 2 x - 3 x}{3 a + x - 2 x}, (a \neq = 0) .$

Evaluate: $lim_{h \to 0} \frac{( a + h ) ^{2} s i n ( a + h ) - a ^{2} s i n a}{h}$

For $x \in R, lim_{x \to \infty} (\frac{x - 3}{x + 2})^{x} =$

Find $lim_{x \to 0} {tan (π /4 + x)}^{1/ x} .$

$lim_{x \to 0} (\frac{1 + 5 x ^{2}}{1 + 3 x ^{2}})^{1/ x^{2}} = ...$

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$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to - \infty} [\frac{x ^{4} s i n ( 1/ x ) + x ^{2}}{( 1 + ∣ x ∣ ^{3} )}] = .........$

If $f (x) = \frac{x - s i n x}{x + c o s ^{2} x}$ , then $lim_{x \to \infty} f (x)$ is

$lim_{x \to \frac{π}{2}} \frac{1 - t a n ( x /2 )}{1 + t a n ( x /2 )} \frac{[ 1 - s i n x ]}{[ π - 2 x ] ^{3}}$ is