Lim x - 0-Incos x-1-x2-1- is equal to

The correct option is A 2 x→ 0lim [ In cos x/√(1+x^2) 1 ]=x→ 0lim=4x→ 0limcos x 1/x^2 [ [∴ In[1+(cos x 1)] ≈(cos x 1)and[√(1+x^2) 1] ≈x^2/4 ] = 1x→ 0limx^2/ ...

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Standard XI

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

lim x → 0[Inc...

Question

$l i m x \to 0 [\frac{I n c o s x}{\sqrt[4]{1 + x^{2}} - 1}] is equal to$

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Solution

The correct option is B
-2

$l i m x \to 0 [\frac{I n c o s x}{\sqrt[4]{1 + x^{2}} - 1}] = l i m x \to 0 = 4 l i m x \to 0 \frac{c o s x - 1}{x^{2}} [[∴ I n [1 + (c o s x - 1)] \approx (c o s x - 1) a n d [\sqrt[4]{1 + x^{2}} - 1] \approx \frac{x^{2}}{4}] = - 1 l i m x \to 0 \frac{x^{2} / 2}{x^{2}} [∵ (1 - c o s x) \approx \frac{x^{2}}{2}] = - 2$

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limx→ 0[In cos x4√1+x2−1]is equal to

The correct option is B -2 limx→ 0[In cos x4√1+x2−1]=limx→ 0=4limx→ 0cos x−1x2[[∴ In[1+(cos x−1)]≈(cos x−1)and[4√1+x2−1]≈x24]=−1limx→ 0x2/2x2 [∵(1−cosx)≈x22]=−2

$l i m x \to 0 [\frac{I n c o s x}{\sqrt[4]{1 + x^{2}} - 1}] is equal to$